![Quiz 1.3
There are exactly 50 equally likely outcomes: s51 through s100. Each of these outcomes
has probability 0.02.
(1) P[{s79}] = 0.02
(2) P[{s100}] = 0.02
(3) P[A] = P[{s90, . . . , s100}] = 11 × 0.02 = 0.22
(4) P[F] = P[{s51, . . . , s59}] = 9 × 0.02 = 0.18
(5) P[T ≥ 80] = P[{s80, . . . , s100}] = 21 × 0.02 = 0.42
(6) P[T < 90] = P[{s51, s52, . . . , s89}] = 39 × 0.02 = 0.78
(7) P[a C grade or better] = P[{s70, . . . , s100}] = 31 × 0.02 = 0.62
(8) P[student passes] = P[{s60, . . . , s100}] = 41 × 0.02 = 0.82
Quiz 1.4
We can describe this experiment by the event space consisting of the four possible
events V B, V L, DB, and DL. We represent these events in the table:
V D
L 0.35 ?
B ? ?
In a roundabout way, the problem statement tells us how to fill in the table. In particular,
P [V ] = 0.7 = P [V L] + P [V B] (1)
P [L] = 0.6 = P [V L] + P [DL] (2)
Since P[V L] = 0.35, we can conclude that P[V B] = 0.35 and that P[DL] = 0.6 −
0.35 = 0.25. This allows us to fill in two more table entries:
V D
L 0.35 0.25
B 0.35 ?
The remaining table entry is filled in by observing that the probabilities must sum to 1.
This implies P[DB] = 0.05 and the complete table is
V D
L 0.35 0.25
B 0.35 0.05
Finding the various probabilities is now straightforward:
3](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-3-320.jpg)
![(1) P[DL] = 0.25
(2) P[D ∪ L] = P[V L] + P[DL] + P[DB] = 0.35 + 0.25 + 0.05 = 0.65.
(3) P[V B] = 0.35
(4) P[V ∪ L] = P[V ] + P[L] − P[V L] = 0.7 + 0.6 − 0.35 = 0.95
(5) P[V ∪ D] = P[S] = 1
(6) P[L B] = P[LLc] = 0
Quiz 1.5
(1) The probability of exactly two voice calls is
P [NV = 2] = P [{vvd, vdv, dvv}] = 0.3 (1)
(2) The probability of at least one voice call is
P [NV ≥ 1] = P [{vdd, dvd, ddv, vvd, vdv, dvv, vvv}] (2)
= 6(0.1) + 0.2 = 0.8 (3)
An easier way to get the same answer is to observe that
P [NV ≥ 1] = 1 − P [NV < 1] = 1 − P [NV = 0] = 1 − P [{ddd}] = 0.8 (4)
(3) The conditional probability of two voice calls followed by a data call given that there
were two voice calls is
P [{vvd} |NV = 2] =
P [{vvd} , NV = 2]
P [NV = 2]
=
P [{vvd}]
P [NV = 2]
=
0.1
0.3
=
1
3
(5)
(4) The conditional probability of two data calls followed by a voice call given there
were two voice calls is
P [{ddv} |NV = 2] =
P [{ddv} , NV = 2]
P [NV = 2]
= 0 (6)
The joint event of the outcome ddv and exactly two voice calls has probability zero
since there is only one voice call in the outcome ddv.
(5) The conditional probability of exactly two voice calls given at least one voice call is
P [NV = 2|Nv ≥ 1] =
P [NV = 2, NV ≥ 1]
P [NV ≥ 1]
=
P [NV = 2]
P [NV ≥ 1]
=
0.3
0.8
=
3
8
(7)
(6) The conditional probability of at least one voice call given there were exactly two
voice calls is
P [NV ≥ 1|NV = 2] =
P [NV ≥ 1, NV = 2]
P [NV = 2]
=
P [NV = 2]
P [NV = 2]
= 1 (8)
Given that there were two voice calls, there must have been at least one voice call.
4](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-4-320.jpg)
![Quiz 1.6
In this experiment, there are four outcomes with probabilities
P[{vv}] = (0.8)2
= 0.64 P[{vd}] = (0.8)(0.2) = 0.16
P[{dv}] = (0.2)(0.8) = 0.16 P[{dd}] = (0.2)2
= 0.04
When checking the independence of any two events A and B, it’s wise to avoid intuition
and simply check whether P[AB] = P[A]P[B]. Using the probabilities of the outcomes,
we now can test for the independence of events.
(1) First, we calculate the probability of the joint event:
P [NV = 2, NV ≥ 1] = P [NV = 2] = P [{vv}] = 0.64 (1)
Next, we observe that
P [NV ≥ 1] = P [{vd, dv, vv}] = 0.96 (2)
Finally, we make the comparison
P [NV = 2] P [NV ≥ 1] = (0.64)(0.96) = P [NV = 2, NV ≥ 1] (3)
which shows the two events are dependent.
(2) The probability of the joint event is
P [NV ≥ 1, C1 = v] = P [{vd, vv}] = 0.80 (4)
From part (a), P[NV ≥ 1] = 0.96. Further, P[C1 = v] = 0.8 so that
P [NV ≥ 1] P [C1 = v] = (0.96)(0.8) = 0.768 = P [NV ≥ 1, C1 = v] (5)
Hence, the events are dependent.
(3) The problem statement that the calls were independent implies that the events the
second call is a voice call, {C2 = v}, and the first call is a data call, {C1 = d} are
independent events. Just to be sure, we can do the calculations to check:
P [C1 = d, C2 = v] = P [{dv}] = 0.16 (6)
Since P[C1 = d]P[C2 = v] = (0.2)(0.8) = 0.16, we confirm that the events are
independent. Note that this shouldn’t be surprising since we used the information that
the calls were independent in the problem statement to determine the probabilities of
the outcomes.
(4) The probability of the joint event is
P [C2 = v, NV is even] = P [{vv}] = 0.64 (7)
Also, each event has probability
P [C2 = v] = P [{dv, vv}] = 0.8, P [NV is even] = P [{dd, vv}] = 0.68 (8)
Thus, P[C2 = v]P[NV is even] = (0.8)(0.68) = 0.544. Since P[C2 = v, NV is even] =
0.544, the events are dependent.
5](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-5-320.jpg)
![Quiz 1.7
Let Fi denote the event that that the user is found on page i. The tree for the experiment
is
¨¨¨¨¨¨ F10.8
Fc
10.2
¨¨¨¨¨¨ F20.8
Fc
20.2
¨¨¨¨¨¨ F30.8
Fc
30.2
The user is found unless all three paging attempts fail. Thus the probability the user is
found is
P [F] = 1 − P Fc
1 Fc
2 Fc
3 = 1 − (0.2)3
= 0.992 (1)
Quiz 1.8
(1) We can view choosing each bit in the code word as a subexperiment. Each subex-
periment has two possible outcomes: 0 and 1. Thus by the fundamental principle of
counting, there are 2 × 2 × 2 × 2 = 24 = 16 possible code words.
(2) An experiment that can yield all possible code words with two zeroes is to choose
which 2 bits (out of 4 bits) will be zero. The other two bits then must be ones. There
are 4
2 = 6 ways to do this. Hence, there are six code words with exactly two zeroes.
For this problem, it is also possible to simply enumerate the six code words:
1100, 1010, 1001, 0101, 0110, 0011.
(3) When the first bit must be a zero, then the first subexperiment of choosing the first
bit has only one outcome. For each of the next three bits, we have two choices. In
this case, there are 1 × 2 × 2 × 2 = 8 ways of choosing a code word.
(4) For the constant ratio code, we can specify a code word by choosing M of the bits to
be ones. The other N − M bits will be zeroes. The number of ways of choosing such
a code word is N
M . For N = 8 and M = 3, there are 8
3 = 56 code words.
Quiz 1.9
(1) In this problem, k bits received in error is the same as k failures in 100 trials. The
failure probability is = 1 − p and the success probability is 1 − = p. That is, the
probability of k bits in error and 100 − k correctly received bits is
P Sk,100−k =
100
k
k
(1 − )100−k
(1)
6](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-6-320.jpg)
![For = 0.01,
P S0,100 = (1 − )100
= (0.99)100
= 0.3660 (2)
P S1,99 = 100(0.01)(0.99)99
= 0.3700 (3)
P S2,98 = 4950(0.01)2
(0.99)9
8 = 0.1849 (4)
P S3,97 = 161, 700(0.01)3
(0.99)97
= 0.0610 (5)
(2) The probability a packet is decoded correctly is just
P [C] = P S0,100 + P S1,99 + P S2,98 + P S3,97 = 0.9819 (6)
Quiz 1.10
Since the chip works only if all n transistors work, the transistors in the chip are like
devices in series. The probability that a chip works is P[C] = pn.
The module works if either 8 chips work or 9 chips work. Let Ck denote the event that
exactly k chips work. Since transistor failures are independent of each other, chip failures
are also independent. Thus each P[Ck] has the binomial probability
P [C8] =
9
8
(P [C])8
(1 − P [C])9−8
= 9p8n
(1 − pn
), (1)
P [C9] = (P [C])9
= p9n
. (2)
The probability a memory module works is
P [M] = P [C8] + P [C9] = p8n
(9 − 8pn
) (3)
Quiz 1.11
R=rand(1,100);
X=(R<= 0.4) ...
+ (2*(R>0.4).*(R<=0.9)) ...
+ (3*(R>0.9));
Y=hist(X,1:3)
For a MATLAB simulation, we first gen-
erate a vector R of 100 random numbers.
Second, we generate vector X as a func-
tion of R to represent the 3 possible out-
comes of a flip. That is, X(i)=1 if flip i
was heads, X(i)=2 if flip i was tails, and
X(i)=3) is flip i landed on the edge.
To see how this works, we note there are three cases:
• If R(i) <= 0.4, then X(i)=1.
• If 0.4 < R(i) and R(i)<=0.9, then X(i)=2.
• If 0.9 < R(i), then X(i)=3.
These three cases will have probabilities 0.4, 0.5 and 0.1. Lastly, we use the hist function
to count how many occurences of each possible value of X(i).
7](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-7-320.jpg)
![Quiz Solutions – Chapter 2
Quiz 2.1
The sample space, probabilities and corresponding grades for the experiment are
Outcome P[·] G
BB 0.36 3.0
BC 0.24 2.5
C B 0.24 2.5
CC 0.16 2
Quiz 2.2
(1) To find c, we recall that the PMF must sum to 1. That is,
3
n=1
PN (n) = c 1 +
1
2
+
1
3
= 1 (1)
This implies c = 6/11. Now that we have found c, the remaining parts are straight-
forward.
(2) P[N = 1] = PN (1) = c = 6/11
(3) P[N ≥ 2] = PN (2) + PN (3) = c/2 + c/3 = 5/11
(4) P[N > 3] = ∞
n=4 PN (n) = 0
Quiz 2.3
Decoding each transmitted bit is an independent trial where we call a bit error a “suc-
cess.” Each bit is in error, that is, the trial is a success, with probability p. Now we can
interpret each experiment in the generic context of independent trials.
(1) The random variable X is the number of trials up to and including the first success.
Similar to Example 2.11, X has the geometric PMF
PX (x) =
p(1 − p)x−1 x = 1, 2, . . .
0 otherwise
(1)
(2) If p = 0.1, then the probability exactly 10 bits are sent is
P [X = 10] = PX (10) = (0.1)(0.9)9
= 0.0387 (2)
8](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-8-320.jpg)
![The probability that at least 10 bits are sent is P[X ≥ 10] = ∞
x=10 PX (x). This
sum is not too hard to calculate. However, its even easier to observe that X ≥ 10 if
the first 10 bits are transmitted correctly. That is,
P [X ≥ 10] = P [first 10 bits are correct] = (1 − p)10
(3)
For p = 0.1, P[X ≥ 10] = 0.910 = 0.3487.
(3) The random variable Y is the number of successes in 100 independent trials. Just as
in Example 2.13, Y has the binomial PMF
PY (y) =
100
y
py
(1 − p)100−y
(4)
If p = 0.01, the probability of exactly 2 errors is
P [Y = 2] = PY (2) =
100
2
(0.01)2
(0.99)98
= 0.1849 (5)
(4) The probability of no more than 2 errors is
P [Y ≤ 2] = PY (0) + PY (1) + PY (2) (6)
= (0.99)100
+ 100(0.01)(0.99)99
+
100
2
(0.01)2
(0.99)98
(7)
= 0.9207 (8)
(5) Random variable Z is the number of trials up to and including the third success. Thus
Z has the Pascal PMF (see Example 2.15)
PZ (z) =
z − 1
2
p3
(1 − p)z−3
(9)
Note that PZ (z) > 0 for z = 3, 4, 5, . . ..
(6) If p = 0.25, the probability that the third error occurs on bit 12 is
PZ (12) =
11
2
(0.25)3
(0.75)9
= 0.0645 (10)
Quiz 2.4
Each of these probabilities can be read off the CDF FY (y). However, we must keep in
mind that when FY (y) has a discontinuity at y0, FY (y) takes the upper value FY (y+
0 ).
(1) P[Y < 1] = FY (1−) = 0
9](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-9-320.jpg)
![(2) P[Y ≤ 1] = FY (1) = 0.6
(3) P[Y > 2] = 1 − P[Y ≤ 2] = 1 − FY (2) = 1 − 0.8 = 0.2
(4) P[Y ≥ 2] = 1 − P[Y < 2] = 1 − FY (2−) = 1 − 0.6 = 0.4
(5) P[Y = 1] = P[Y ≤ 1] − P[Y < 1] = FY (1+) − FY (1−) = 0.6
(6) P[Y = 3] = P[Y ≤ 3] − P[Y < 3] = FY (3+) − FY (3−) = 0.8 − 0.8 = 0
Quiz 2.5
(1) With probability 0.7, a call is a voice call and C = 25. Otherwise, with probability
0.3, we have a data call and C = 40. This corresponds to the PMF
PC (c) =
⎧
⎨
⎩
0.7 c = 25
0.3 c = 40
0 otherwise
(1)
(2) The expected value of C is
E [C] = 25(0.7) + 40(0.3) = 29.5 cents (2)
Quiz 2.6
(1) As a function of N, the cost T is
T = 25N + 40(3 − N) = 120 − 15N (1)
(2) To find the PMF of T , we can draw the following tree:
¨¨¨
¨¨¨¨N=0
0.1
rrrrr
rrN=3
0.3
$$$$$$$N=10.3
ˆˆˆˆˆˆˆN=20.3
•T =120
•T =105
•T =90
•T =75
From the tree, we can write down the PMF of T :
PT (t) =
⎧
⎨
⎩
0.3 t = 75, 90, 105
0.1 t = 120
0 otherwise
(2)
From the PMF PT (t), the expected value of T is
E [T ] = 75PT (75) + 90PT (90) + 105PT (105) + 120PT (120) (3)
= (75 + 90 + 105)(0.3) + 120(0.1) = 62 (4)
10](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-10-320.jpg)
![Quiz 2.7
(1) Using Definition 2.14, the expected number of applications is
E [A] =
4
a=1
aPA (a) = 1(0.4) + 2(0.3) + 3(0.2) + 4(0.1) = 2 (1)
(2) The number of memory chips is M = g(A) where
g(A) =
⎧
⎨
⎩
4 A = 1, 2
6 A = 3
8 A = 4
(2)
(3) By Theorem 2.10, the expected number of memory chips is
E [M] =
4
a=1
g(A)PA (a) = 4(0.4) + 4(0.3) + 6(0.2) + 8(0.1) = 4.8 (3)
Since E[A] = 2, g(E[A]) = g(2) = 4. However, E[M] = 4.8 = g(E[A]). The two
quantities are different because g(A) is not of the form αA + β.
Quiz 2.8
The PMF PN (n) allows to calculate each of the desired quantities.
(1) The expected value of N is
E [N] =
2
n=0
nPN (n) = 0(0.1) + 1(0.4) + 2(0.5) = 1.4 (1)
(2) The second moment of N is
E N2
=
2
n=0
n2
PN (n) = 02
(0.1) + 12
(0.4) + 22
(0.5) = 2.4 (2)
(3) The variance of N is
Var[N] = E N2
− (E [N])2
= 2.4 − (1.4)2
= 0.44 (3)
(4) The standard deviation is σN =
√
Var[N] =
√
0.44 = 0.663.
11](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-11-320.jpg)
![Quiz 2.9
(1) From the problem statement, we learn that the conditional PMF of N given the event
I is
PN|I (n) =
0.02 n = 1, 2, . . . , 50
0 otherwise
(1)
(2) Also from the problem statement, the conditional PMF of N given the event T is
PN|T (n) =
0.2 n = 1, 2, 3, 4, 5
0 otherwise
(2)
(3) The problem statement tells us that P[T ] = 1 − P[I] = 3/4. From Theorem 1.10
(the law of total probability), we find the PMF of N is
PN (n) = PN|T (n) P [T ] + PN|I (n) P [I] (3)
=
⎧
⎨
⎩
0.2(0.75) + 0.02(0.25) n = 1, 2, 3, 4, 5
0(0.75) + 0.02(0.25) n = 6, 7, . . . , 50
0 otherwise
(4)
=
⎧
⎨
⎩
0.155 n = 1, 2, 3, 4, 5
0.005 n = 6, 7, . . . , 50
0 otherwise
(5)
(4) First we find
P [N ≤ 10] =
10
n=1
PN (n) = (0.155)(5) + (0.005)(5) = 0.80 (6)
By Theorem 2.17, the conditional PMF of N given N ≤ 10 is
PN|N≤10 (n) =
PN (n)
P[N≤10] n ≤ 10
0 otherwise
(7)
=
⎧
⎨
⎩
0.155/0.8 n = 1, 2, 3, 4, 5
0.005/0.8 n = 6, 7, 8, 9, 10
0 otherwise
(8)
=
⎧
⎨
⎩
0.19375 n = 1, 2, 3, 4, 5
0.00625 n = 6, 7, 8, 9, 10
0 otherwise
(9)
(5) Once we have the conditional PMF, calculating conditional expectations is easy.
E [N|N ≤ 10] =
n
nPN|N≤10 (n) (10)
=
5
n=1
n(0.19375) +
10
n=6
n(0.00625) (11)
= 3.15625 (12)
12](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-12-320.jpg)
![0 50 100
0
2
4
6
8
10
0 500 1000
0
2
4
6
8
10
(a) samplemean(100) (b) samplemean(1000)
Figure 1: Two examples of the output of samplemean(k)
(6) To find the conditional variance, we first find the conditional second moment
E N2
|N ≤ 10 =
n
n2
PN|N≤10 (n) (13)
=
5
n=1
n2
(0.19375) +
10
n=6
n2
(0.00625) (14)
= 55(0.19375) + 330(0.00625) = 12.71875 (15)
The conditional variance is
Var[N|N ≤ 10] = E N2
|N ≤ 10 − (E [N|N ≤ 10])2
(16)
= 12.71875 − (3.15625)2
= 2.75684 (17)
Quiz 2.10
The function samplemean(k) generates and plots five mn sequences for n = 1, 2, . . . , k.
The ith column M(:,i) of M holds a sequence m1, m2, . . . , mk.
function M=samplemean(k);
K=(1:k)’;
M=zeros(k,5);
for i=1:5,
X=duniformrv(0,10,k);
M(:,i)=cumsum(X)./K;
end;
plot(K,M);
Examples of the function calls (a) samplemean(100) and (b) samplemean(1000)
are shown in Figure 1. Each time samplemean(k) is called produces a random output.
What is observed in these figures is that for small n, mn is fairly random but as n gets
13](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-13-320.jpg)
![large, mn gets close to E[X] = 5. Although each sequence m1, m2, . . . that we generate is
random, the sequences always converges to E[X]. This random convergence is analyzed
in Chapter 7.
14](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-14-320.jpg)
![Quiz Solutions – Chapter 3
Quiz 3.1
The CDF of Y is
0 2 4
0
0.5
1
y
F
Y
(y)
FY (y) =
⎧
⎨
⎩
0 y < 0
y/4 0 ≤ y ≤ 4
1 y > 4
(1)
From the CDF FY (y), we can calculate the probabilities:
(1) P[Y ≤ −1] = FY (−1) = 0
(2) P[Y ≤ 1] = FY (1) = 1/4
(3) P[2 < Y ≤ 3] = FY (3) − FY (2) = 3/4 − 2/4 = 1/4
(4) P[Y > 1.5] = 1 − P[Y ≤ 1.5] = 1 − FY (1.5) = 1 − (1.5)/4 = 5/8
Quiz 3.2
(1) First we will find the constant c and then we will sketch the PDF. To find c, we use
the fact that
∞
−∞ fX (x) dx = 1. We will evaluate this integral using integration by
parts:
∞
−∞
fX (x) dx =
∞
0
cxe−x/2
dx (1)
= −2cxe−x/2
∞
0
=0
+
∞
0
2ce−x/2
dx (2)
= −4ce−x/2
∞
0
= 4c (3)
Thus c = 1/4 and X has the Erlang (n = 2, λ = 1/2) PDF
0 5 10 15
0
0.1
0.2
x
f
X
(x)
fX (x) =
(x/4)e−x/2 x ≥ 0
0 otherwise
(4)
15](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-15-320.jpg)
![(2) To find the CDF FX (x), we first note X is a nonnegative random variable so that
FX (x) = 0 for all x < 0. For x ≥ 0,
FX (x) =
x
0
fX (y) dy =
x
0
y
4
e−y/2
dy (5)
= −
y
2
e−y/2
x
0
−
x
0
−
1
2
e−y/2
dy (6)
= 1 −
x
2
e−x/2
− e−x/2
(7)
The complete expression for the CDF is
0 5 10 15
0
0.5
1
x
F
X
(x)
FX (x) =
1 − x
2 + 1 e−x/2 x ≥ 0
0 otherwise
(8)
(3) From the CDF FX (x),
P [0 ≤ X ≤ 4] = FX (4) − FX (0) = 1 − 3e−2
. (9)
(4) Similarly,
P [−2 ≤ X ≤ 2] = FX (2) − FX (−2) = 1 − 3e−1
. (10)
Quiz 3.3
The PDF of Y is
−2 0 2
0
1
2
3
y
f
Y
(y)
fY (y) =
3y2/2 −1 ≤ y ≤ 1,
0 otherwise.
(1)
(1) The expected value of Y is
E [Y] =
∞
−∞
y fY (y) dy =
1
−1
(3/2)y3
dy = (3/8)y4
1
−1
= 0. (2)
Note that the above calculation wasn’t really necessary because E[Y] = 0 whenever
the PDF fY (y) is an even function (i.e., fY (y) = fY (−y)).
(2) The second moment of Y is
E Y2
=
∞
−∞
y2
fY (y) dy =
1
−1
(3/2)y4
dy = (3/10)y5
1
−1
= 3/5. (3)
16](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-16-320.jpg)
![(3) The variance of Y is
Var[Y] = E Y2
− (E [Y])2
= 3/5. (4)
(4) The standard deviation of Y is σY =
√
Var[Y] =
√
3/5.
Quiz 3.4
(1) When X is an exponential (λ) random variable, E[X] = 1/λ and Var[X] = 1/λ2.
Since E[X] = 3 and Var[X] = 9, we must have λ = 1/3. The PDF of X is
fX (x) =
(1/3)e−x/3 x ≥ 0,
0 otherwise.
(1)
(2) We know X is a uniform (a, b) random variable. To find a and b, we apply Theo-
rem 3.6 to write
E [X] =
a + b
2
= 3 Var[X] =
(b − a)2
12
= 9. (2)
This implies
a + b = 6, b − a = ±6
√
3. (3)
The only valid solution with a < b is
a = 3 − 3
√
3, b = 3 + 3
√
3. (4)
The complete expression for the PDF of X is
fX (x) =
1/(6
√
3) 3 − 3
√
3 ≤ x < 3 + 3
√
3,
0 otherwise.
(5)
Quiz 3.5
Each of the requested probabilities can be calculated using (z) function and Table 3.1
or Q(z) and Table 3.2. We start with the sketches.
(1) The PDFs of X and Y are shown below. The fact that Y has twice the standard
deviation of X is reflected in the greater spread of fY (y). However, it is important
to remember that as the standard deviation increases, the peak value of the Gaussian
PDF goes down.
−5 0 5
0
0.2
0.4
x y
f
X
(x)f
Y
(y)
← f
X
(x)
← f
Y
(y)
17](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-17-320.jpg)
![(2) Since X is Gaussian (0, 1),
P [−1 < X ≤ 1] = FX (1) − FX (−1) (1)
= (1) − (−1) = 2 (1) − 1 = 0.6826. (2)
(3) Since Y is Gaussian (0, 2),
P [−1 < Y ≤ 1] = FY (1) − FY (−1) (3)
=
1
σY
−
−1
σY
= 2
1
2
− 1 = 0.383. (4)
(4) Again, since X is Gaussian (0, 1), P[X > 3.5] = Q(3.5) = 2.33 × 10−4.
(5) Since Y is Gaussian (0, 2), P[Y > 3.5] = Q(3.5
2 ) = Q(1.75) = 1 − (1.75) =
0.0401.
Quiz 3.6
The CDF of X is
−2 0 2
0
0.5
1
x
F
X
(x)
FX (x) =
⎧
⎨
⎩
0 x < −1,
(x + 1)/4 −1 ≤ x < 1,
1 x ≥ 1.
(1)
The following probabilities can be read directly from the CDF:
(1) P[X ≤ 1] = FX (1) = 1.
(2) P[X < 1] = FX (1−) = 1/2.
(3) P[X = 1] = FX (1+) − FX (1−) = 1 − 1/2 = 1/2.
(4) We find the PDF fY (y) by taking the derivative of FY (y). The resulting PDF is
−2 0 2
0
0.5
x
f
X
(x)
0.5
fX (x) =
⎧
⎨
⎩
1/4 −1 ≤ x < 1,
(1/2)δ(x − 1) x = 1,
0 otherwise.
(2)
Quiz 3.7
18](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-18-320.jpg)
![(1) Since X is always nonnegative, FX (x) = 0 for x < 0. Also, FX (x) = 1 for x ≥ 2
since its always true that x ≤ 2. Lastly, for 0 ≤ x ≤ 2,
FX (x) =
x
−∞
fX (y) dy =
x
0
(1 − y/2) dy = x − x2
/4. (1)
The complete CDF of X is
−1 0 1 2 3
0
0.5
1
x
F
X
(x)
FX (x) =
⎧
⎨
⎩
0 x < 0,
x − x2/4 0 ≤ x ≤ 2,
1 x > 2.
(2)
(2) The probability that Y = 1 is
P [Y = 1] = P [X ≥ 1] = 1 − FX (1) = 1 − 3/4 = 1/4. (3)
(3) Since X is nonnegative, Y is also nonnegative. Thus FY (y) = 0 for y < 0. Also,
because Y ≤ 1, FY (y) = 1 for all y ≥ 1. Finally, for 0 < y < 1,
FY (y) = P [Y ≤ y] = P [X ≤ y] = FX (y) . (4)
Using the CDF FX (x), the complete expression for the CDF of Y is
−1 0 1 2 3
0
0.5
1
y
F
Y
(y)
FY (y) =
⎧
⎨
⎩
0 y < 0,
y − y2/4 0 ≤ y < 1,
1 y ≥ 1.
(5)
As expected, we see that the jump in FY (y) at y = 1 is exactly equal to P[Y = 1].
(4) By taking the derivative of FY (y), we obtain the PDF fY (y). Note that when y < 0
or y > 1, the PDF is zero.
−1 0 1 2 3
0
0.5
1
1.5
y
f
Y
(y)
0.25
fY (y) =
1 − y/2 + (1/4)δ(y − 1) 0 ≤ y ≤ 1
0 otherwise
(6)
Quiz 3.8
(1) P[Y ≤ 6] =
6
−∞ fY (y) dy =
6
0 (1/10) dy = 0.6 .
19](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-19-320.jpg)
![(2) From Definition 3.15, the conditional PDF of Y given Y ≤ 6 is
fY|Y≤6 (y) =
fY (y)
P[Y≤6] y ≤ 6,
0 otherwise,
=
1/6 0 ≤ y ≤ 6,
0 otherwise.
(1)
(3) The probability Y > 8 is
P [Y > 8] =
10
8
1
10
dy = 0.2 . (2)
(4) From Definition 3.15, the conditional PDF of Y given Y > 8 is
fY|Y>8 (y) =
fY (y)
P[Y>8] y > 8,
0 otherwise,
=
1/2 8 < y ≤ 10,
0 otherwise.
(3)
(5) From the conditional PDF fY|Y≤6(y), we can calculate the conditional expectation
E [Y|Y ≤ 6] =
∞
−∞
y fY|Y≤6 (y) dy =
6
0
y
6
dy = 3. (4)
(6) From the conditional PDF fY|Y>8(y), we can calculate the conditional expectation
E [Y|Y > 8] =
∞
−∞
yfY|Y>8 (y) dy =
10
8
y
2
dy = 9. (5)
Quiz 3.9
A natural way to produce random variables with PDF fT|T >2(t) is to generate samples
of T with PDF fT (t) and then to discard those samples which fail to satisfy the condition
T > 2. Here is a MATLAB function that uses this method:
function t=t2rv(m)
i=0;lambda=1/3;
t=zeros(m,1);
while (i<m),
x=exponentialrv(lambda,1);
if (x>2)
t(i+1)=x;
i=i+1;
end
end
A second method exploits the fact that if T is an exponential (λ) random variable, then
T = T + 2 has PDF fT (t) = fT|T >2(t). In this case the command
t=2.0+exponentialrv(1/3,m)
generates the vector t.
20](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-20-320.jpg)
![Quiz Solutions – Chapter 4
Quiz 4.1
Each value of the joint CDF can be found by considering the corresponding probability.
(1) FX,Y (−∞, 2) = P[X ≤ −∞, Y ≤ 2] ≤ P[X ≤ −∞] = 0 since X cannot take on
the value −∞.
(2) FX,Y (∞, ∞) = P[X ≤ ∞, Y ≤ ∞] = 1. This result is given in Theorem 4.1.
(3) FX,Y (∞, y) = P[X ≤ ∞, Y ≤ y] = P[Y ≤ y] = FY (y).
(4) FX,Y (∞, −∞) = P[X ≤ ∞, Y ≤ −∞] = 0 since Y cannot take on the value −∞.
Quiz 4.2
From the joint PMF of Q and G given in the table, we can calculate the requested
probabilities by summing the PMF over those values of Q and G that correspond to the
event.
(1) The probability that Q = 0 is
P [Q = 0] = PQ,G (0, 0) + PQ,G (0, 1) + PQ,G (0, 2) + PQ,G (0, 3) (1)
= 0.06 + 0.18 + 0.24 + 0.12 = 0.6 (2)
(2) The probability that Q = G is
P [Q = G] = PQ,G (0, 0) + PQ,G (1, 1) = 0.18 (3)
(3) The probability that G > 1 is
P [G > 1] =
3
g=2
1
q=0
PQ,G (q, g) (4)
= 0.24 + 0.16 + 0.12 + 0.08 = 0.6 (5)
(4) The probability that G > Q is
P [G > Q] =
1
q=0
3
g=q+1
PQ,G (q, g) (6)
= 0.18 + 0.24 + 0.12 + 0.16 + 0.08 = 0.78 (7)
21](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-21-320.jpg)
![Quiz 4.3
By Theorem 4.3, the marginal PMF of H is
PH (h) =
b=0,2,4
PH,B (h, b) (1)
For each value of h, this corresponds to calculating the row sum across the table of the joint
PMF. Similarly, the marginal PMF of B is
PB (b) =
1
h=−1
PH,B (h, b) (2)
For each value of b, this corresponds to the column sum down the table of the joint PMF.
The easiest way to calculate these marginal PMFs is to simply sum each row and column:
PH,B (h, b) b = 0 b = 2 b = 4 PH (h)
h = −1 0 0.4 0.2 0.6
h = 0 0.1 0 0.1 0.2
h = 1 0.1 0.1 0 0.2
PB (b) 0.2 0.5 0.3
(3)
Quiz 4.4
To find the constant c, we apply
∞
−∞
∞
−∞ fX,Y (x, y) dx dy = 1. Specifically,
∞
−∞
∞
−∞
fX,Y (x, y) dx dy =
2
0
1
0
cxy dx dy (1)
= c
2
0
y x2
/2
1
0
dy (2)
= (c/2)
2
0
y dy = (c/4)y2
2
0
= c (3)
Thus c = 1. To calculate P[A], we write
P [A] =
A
fX,Y (x, y) dx dy (4)
To integrate over A, we convert to polar coordinates using the substitutions x = r cos θ,
y = r sin θ and dx dy = r dr dθ, yielding
Y
X
1
1
2
A
P [A] =
π/2
0
1
0
r2
sin θ cos θ r dr dθ (5)
=
1
0
r3
dr
π/2
0
sin θ cos θ dθ (6)
= r4
/4
1
0
⎛
⎝ sin2
θ
2
π/2
0
⎞
⎠ = 1/8 (7)
22](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-22-320.jpg)
![(B) First, we observe that since 0 ≤ X ≤ 1 and 0 ≤ Y ≤ 1, W = XY satisfies
0 ≤ W ≤ 1. Thus fW (0) = 0 and fW (1) = 1. For 0 < w < 1, we calculate the
CDF FW (w) = P[W ≤ w]. As shown below, integrating over the region W ≤ w
is fairly complex. The calculus is simpler if we integrate over the region XY > w.
Specifically,
Y
X
1
1
XY > w
w
w XY = w
FW (w) = 1 − P [XY > w] (2)
= 1 −
1
w
1
w/x
dy dx (3)
= 1 −
1
w
(1 − w/x) dx (4)
= 1 − x − w ln x|x=1
x=w (5)
= 1 − (1 − w + w ln w) = w − w ln w (6)
The complete expression for the CDF is
FW (w) =
⎧
⎨
⎩
0 w < 0
w − w ln w 0 ≤ w ≤ 1
1 w > 1
(7)
By taking the derivative of the CDF, we find the PDF is
fW (w) =
d FW (w)
dw
=
⎧
⎨
⎩
0 w < 0
− ln w 0 ≤ w ≤ 1
0 w > 1
(8)
Quiz 4.7
(A) It is helpful to first make a table that includes the marginal PMFs.
PL,T (l, t) t = 40 t = 60 PL(l)
l = 1 0.15 0.1 0.25
l = 2 0.3 0.2 0.5
l = 3 0.15 0.1 0.25
PT (t) 0.6 0.4
(1) The expected value of L is
E [L] = 1(0.25) + 2(0.5) + 3(0.25) = 2. (1)
Since the second moment of L is
E L2
= 12
(0.25) + 22
(0.5) + 32
(0.25) = 4.5, (2)
the variance of L is
Var [L] = E L2
− (E [L])2
= 0.5. (3)
24](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-24-320.jpg)
![(2) The expected value of T is
E [T ] = 40(0.6) + 60(0.4) = 48. (4)
The second moment of T is
E T 2
= 402
(0.6) + 602
(0.4) = 2400. (5)
Thus
Var[T ] = E T 2
− (E [T ])2
= 2400 − 482
= 96. (6)
(3) The correlation is
E [LT ] =
t=40,60
3
l=1
lt PLT (lt) (7)
= 1(40)(0.15) + 2(40)(0.3) + 3(40)(0.15) (8)
+ 1(60)(0.1) + 2(60)(0.2) + 3(60)(0.1) (9)
= 96 (10)
(4) From Theorem 4.16(a), the covariance of L and T is
Cov [L, T ] = E [LT ] − E [L] E [T ] = 96 − 2(48) = 0 (11)
(5) Since Cov[L, T ] = 0, the correlation coefficient is ρL,T = 0.
(B) As in the discrete case, the calculations become easier if we first calculate the marginal
PDFs fX (x) and fY (y). For 0 ≤ x ≤ 1,
fX (x) =
∞
−∞
fX,Y (x, y) dy =
2
0
xy dy =
1
2
xy2
y=2
y=0
= 2x (12)
Similarly, for 0 ≤ y ≤ 2,
fY (y) =
∞
−∞
fX,Y (x, y) dx =
2
0
xy dx =
1
2
x2
y
x=1
x=0
=
y
2
(13)
The complete expressions for the marginal PDFs are
fX (x) =
2x 0 ≤ x ≤ 1
0 otherwise
fY (y) =
y/2 0 ≤ y ≤ 2
0 otherwise
(14)
From the marginal PDFs, it is straightforward to calculate the various expectations.
25](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-25-320.jpg)
![(1) The first and second moments of X are
E [X] =
∞
−∞
x fX (x) dx =
1
0
2x2
dx =
2
3
(15)
E X2
=
∞
−∞
x2
fX (x) dx =
1
0
2x3
dx =
1
2
(16)
(17)
The variance of X is Var[X] = E[X2] − (E[X])2 = 1/18.
(2) The first and second moments of Y are
E [Y] =
∞
−∞
y fY (y) dy =
2
0
1
2
y2
dy =
4
3
(18)
E Y2
=
∞
−∞
y2
fY (y) dy =
2
0
1
2
y3
dy = 2 (19)
The variance of Y is Var[Y] = E[Y2] − (E[Y])2 = 2 − 16/9 = 2/9.
(3) The correlation of X and Y is
E [XY] =
∞
−∞
∞
−∞
xy fX,Y (x, y) dx, dy (20)
=
1
0
2
0
x2
y2
dx, dy =
x3
3
1
0
y3
3
2
0
=
8
9
(21)
(4) The covariance of X and Y is
Cov [X, Y] = E [XY] − E [X] E [Y] =
8
9
−
2
3
4
3
= 0. (22)
(5) Since Cov[X, Y] = 0, the correlation coefficient is ρX,Y = 0.
Quiz 4.8
(A) Since the event V > 80 occurs only for the pairs (L, T ) = (2, 60), (L, T ) = (3, 40)
and (L, T ) = (3, 60),
P [A] = P [V > 80] = PL,T (2, 60) + PL,T (3, 40) + PL,T (3, 60) = 0.45 (1)
By Definition 4.9,
PL,T|A (l, t) =
PL,T (l,t)
P[A] lt > 80
0 otherwise
(2)
26](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-26-320.jpg)
![We can represent this conditional PMF in the following table:
PL,T |A(l, t) t = 40 t = 60
l = 1 0 0
l = 2 0 4/9
l = 3 1/3 2/9
The conditional expectation of V can be found from the conditional PMF.
E [V |A] =
l t
lt PL,T |A (l, t) (3)
= (2 · 60)
4
9
+ (3 · 40)
1
3
+ (3 · 60)
2
9
= 133
1
3
(4)
For the conditional variance Var[V |A], we first find the conditional second moment
E V 2
|A =
l t
(lt)2
PL,T |A (l, t) (5)
= (2 · 60)2 4
9
+ (3 · 40)2 1
3
+ (3 · 60)2 2
9
= 18, 400 (6)
It follows that
Var [V |A] = E V 2
|A − (E [V |A])2
= 622
2
9
(7)
(B) For continuous random variables X and Y, we first calculate the probability of the
conditioning event.
P [B] =
B
fX,Y (x, y) dx dy =
60
40
3
80/y
xy
4000
dx dy (8)
=
60
40
y
4000
x2
2
3
80/y
dy (9)
=
60
40
y
4000
9
2
−
3200
y2
dy (10)
=
9
8
−
4
5
ln
3
2
≈ 0.801 (11)
The conditional PDF of X and Y is
fX,Y|B (x, y) =
fX,Y (x, y) /P [B] (x, y) ∈ B
0 otherwise
(12)
=
K xy 40 ≤ y ≤ 60, 80/y ≤ x ≤ 3
0 otherwise
(13)
27](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-27-320.jpg)
![where K = (4000P[B])−1. The conditional expectation of W given event B is
E [W|B] =
∞
−∞
∞
−∞
xy fX,Y|B (x, y) dx dy (14)
=
60
40
3
80/y
K x2
y2
dx dy (15)
= (K/3)
60
40
y2
x3
x=3
x=80/y
dy (16)
= (K/3)
60
40
27y2
− 803
/y dy (17)
= (K/3) 9y3
− 803
ln y
60
40
≈ 120.78 (18)
The conditional second moment of K given B is
E W2
|B =
∞
−∞
∞
−∞
(xy)2
fX,Y|B (x, y) dx dy (19)
=
60
40
3
80/y
K x3
y3
dx dy (20)
= (K/4)
60
40
y3
x4
x=3
x=80/y
dy (21)
= (K/4)
60
40
81y3
− 804
/y dy (22)
= (K/4) (81/4)y4
− 804
ln y
60
40
≈ 16, 116.10 (23)
It follows that the conditional variance of W given B is
Var [W|B] = E W2
|B − (E [W|B])2
≈ 1528.30 (24)
Quiz 4.9
(A) (1) The joint PMF of A and B can be found from the marginal and conditional
PMFs via PA,B(a, b) = PB|A(b|a)PA(a). Incorporating the information from
the given conditional PMFs can be confusing, however. Consequently, we can
note that A has range SA = {0, 2} and B has range SB = {0, 1}. A table of the
joint PMF will include all four possible combinations of A and B. The general
form of the table is
PA,B(a, b) b = 0 b = 1
a = 0 PB|A(0|0)PA(0) PB|A(1|0)PA(0)
a = 2 PB|A(0|2)PA(2) PB|A(1|2)PA(2)
28](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-28-320.jpg)
![Substituting values from PB|A(b|a) and PA(a), we have
PA,B(a, b) b = 0 b = 1
a = 0 (0.8)(0.4) (0.2)(0.4)
a = 2 (0.5)(0.6) (0.5)(0.6)
or
PA,B(a, b) b = 0 b = 1
a = 0 0.32 0.08
a = 2 0.3 0.3
(2) Given the conditional PMF PB|A(b|2), it is easy to calculate the conditional
expectation
E [B|A = 2] =
1
b=0
bPB|A (b|2) = (0)(0.5) + (1)(0.5) = 0.5 (1)
(3) From the joint PMF PA,B(a, b), we can calculate the the conditional PMF
PA|B (a|0) =
PA,B (a, 0)
PB (0)
=
⎧
⎨
⎩
0.32/0.62 a = 0
0.3/0.62 a = 2
0 otherwise
(2)
=
⎧
⎨
⎩
16/31 a = 0
15/31 a = 2
0 otherwise
(3)
(4) We can calculate the conditional variance Var[A|B = 0] using the conditional
PMF PA|B(a|0). First we calculate the conditional expected value
E [A|B = 0] =
a
aPA|B (a|0) = 0(16/31) + 2(15/31) = 30/31 (4)
The conditional second moment is
E A2
|B = 0 =
a
a2
PA|B (a|0) = 02
(16/31) + 22
(15/31) = 60/31 (5)
The conditional variance is then
Var[A|B = 0] = E A2
|B = 0 − (E [A|B = 0])2
=
960
961
(6)
(B) (1) The joint PDF of X and Y is
fX,Y (x, y) = fY|X (y|x) fX (x) =
6y 0 ≤ y ≤ x, 0 ≤ x ≤ 1
0 otherwise
(7)
(2) From the given conditional PDF fY|X (y|x),
fY|X (y|1/2) =
8y 0 ≤ y ≤ 1/2
0 otherwise
(8)
29](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-29-320.jpg)
![(3) The conditional PDF of Y given X = 1/2 is fX|Y (x|1/2) = fX,Y (x, 1/2)/fY (1/2).
To find fY (1/2), we integrate the joint PDF.
fY (1/2) =
∞
−∞
fX,1/2 ( ) dx =
1
1/2
6(1/2) dx = 3/2 (9)
Thus, for 1/2 ≤ x ≤ 1,
fX|Y (x|1/2) =
fX,Y (x, 1/2)
fY (1/2)
=
6(1/2)
3/2
= 2 (10)
(4) From the pervious part, we see that given Y = 1/2, the conditional PDF of X
is uniform (1/2, 1). Thus, by the definition of the uniform (a, b) PDF,
Var [X|Y = 1/2] =
(1 − 1/2)2
12
=
1
48
(11)
Quiz 4.10
(A) (1) For random variables X and Y from Example 4.1, we observe that PY (1) =
0.09 and PX (0) = 0.01. However,
PX,Y (0, 1) = 0 = PX (0) PY (1) (1)
Since we have found a pair x, y such that PX,Y (x, y) = PX (x)PY (y), we can
conclude that X and Y are dependent. Note that whenever PX,Y (x, y) = 0,
independence requires that either PX (x) = 0 or PY (y) = 0.
(2) For random variables Q and G from Quiz 4.2, it is not obvious whether they
are independent. Unlike X and Y in part (a), there are no obvious pairs q, g
that fail the independence requirement. In this case, we calculate the marginal
PMFs from the table of the joint PMF PQ,G(q, g) in Quiz 4.2.
PQ,G(q, g) g = 0 g = 1 g = 2 g = 3 PQ(q)
q = 0 0.06 0.18 0.24 0.12 0.60
q = 1 0.04 0.12 0.16 0.08 0.40
PG(g) 0.10 0.30 0.40 0.20
Careful study of the table will verify that PQ,G(q, g) = PQ(q)PG(g) for every
pair q, g. Hence Q and G are independent.
(B) (1) Since X1 and X2 are independent,
fX1,X2 (x1, x2) = fX1 (x1) fX2 (x2) (2)
=
(1 − x1/2)(1 − x2/2) 0 ≤ x1 ≤ 2, 0 ≤ x2 ≤ 2
0 otherwise
(3)
30](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-30-320.jpg)
![(2) Let FX (x) denote the CDF of both X1 and X2. The CDF of Z = max(X1, X2)
is found by observing that Z ≤ z iff X1 ≤ z and X2 ≤ z. That is,
P [Z ≤ z] = P [X1 ≤ z, X2 ≤ z] (4)
= P [X1 ≤ z] P [X2 ≤ z] = [FX (z)]2
(5)
To complete the problem, we need to find the CDF of each Xi . From the PDF
fX (x), the CDF is
FX (x) =
x
−∞
fX (y) dy =
⎧
⎨
⎩
0 x < 0
x − x2/4 0 ≤ x ≤ 2
1 x > 2
(6)
Thus for 0 ≤ z ≤ 2,
FZ (z) = (z − z2
/4)2
(7)
The complete expression for the CDF of Z is
FZ (z) =
⎧
⎨
⎩
0 z < 0
(z − z2/4)2 0 ≤ z ≤ 2
1 z > 1
(8)
Quiz 4.11
This problem just requires identifying the various terms in Definition 4.17 and Theo-
rem 4.29. Specifically, from the problem statement, we know that ρ = 1/2,
µ1 = µX = 0, µ2 = µY = 0, (1)
and that
σ1 = σX = 1, σ2 = σY = 1. (2)
(1) Applying these facts to Definition 4.17, we have
fX,Y (x, y) =
1
√
3π2
e−2(x2−xy+y2)/3
. (3)
(2) By Theorem 4.30, the conditional expected value and standard deviation of X given
Y = y are
E [X|Y = y] = y/2 ˜σX = σ2
1 (1 − ρ2
) = 3/4. (4)
When Y = y = 2, we see that E[X|Y = 2] = 1 and Var[X|Y = 2] = 3/4. The
conditional PDF of X given Y = 2 is simply the Gaussian PDF
fX|Y (x|2) =
1
√
3π/2
e−2(x−1)2/3
. (5)
31](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-31-320.jpg)
![Quiz 4.12
One straightforward method is to follow the approach of Example 4.28. Instead, we use
an alternate approach. First we observe that X has the discrete uniform (1, 4) PMF. Also,
given X = x, Y has a discrete uniform (1, x) PMF. That is,
PX (x) =
1/4 x = 1, 2, 3, 4,
0 otherwise,
PY|X (y|x) =
1/x y = 1, . . . , x
0 otherwise
(1)
Given X = x, and an independent uniform (0, 1) random variable U, we can generate a
sample value of Y with a discrete uniform (1, x) PMF via Y = xU . This observation
prompts the following program:
function xy=dtrianglerv(m)
sx=[1;2;3;4];
px=0.25*ones(4,1);
x=finiterv(sx,px,m);
y=ceil(x.*rand(m,1));
xy=[x’;y’];
32](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-32-320.jpg)
![Quiz Solutions – Chapter 5
Quiz 5.1
We find P[C] by integrating the joint PDF over the region of interest. Specifically,
P [C] =
1/2
0
dy2
y2
0
dy1
1/2
0
dy4
y4
0
4dy3 (1)
= 4
1/2
0
y2 dy2
1/2
0
y4 dy4 = 1/4. (2)
Quiz 5.2
By definition of A, Y1 = X1, Y2 = X2 − X1 and Y3 = X3 − X2. Since 0 < X1 < X2 <
X3, each Yi must be a strictly positive integer. Thus, for y1, y2, y3 ∈ {1, 2, . . .},
PY (y) = P [Y1 = y1, Y2 = y2, Y3 = y3] (1)
= P [X1 = y1, X2 − X1 = y2, X3 − X2 = y3] (2)
= P [X1 = y1, X2 = y2 + y1, X3 = y3 + y2 + y1] (3)
= (1 − p)3
py1+y2+y3 (4)
By defining the vector a = 1 1 1 , the complete expression for the joint PMF of Y is
PY (y) =
(1 − p)pa y y1, y2, y3 ∈ {1, 2, . . .}
0 otherwise
(5)
Quiz 5.3
First we note that each marginal PDF is nonzero only if any subset of the xi obeys the
ordering contraints 0 ≤ x1 ≤ x2 ≤ x3 ≤ 1. Within these constraints, we have
fX1,X2 (x1, x2) =
∞
−∞
fX (x) dx3 =
1
x2
6 dx3 = 6(1 − x2), (1)
fX2,X3 (x2, x3) =
∞
−∞
fX (x) dx1 =
x2
0
6 dx1 = 6x2, (2)
fX1,X3 (x1, x3) =
∞
−∞
fX (x) dx2 =
x3
x1
6 dx2 = 6(x3 − x1). (3)
In particular, we must keep in mind that fX1,X2(x1, x2) = 0 unless 0 ≤ x1 ≤ x2 ≤ 1,
fX2,X3(x2, x3) = 0 unless 0 ≤ x2 ≤ x3 ≤ 1, and that fX1,X3(x1, x3) = 0 unless 0 ≤ x1 ≤
33](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-33-320.jpg)
![We must verify that V and W are independent. For 0 ≤ v1 ≤ v2 ≤ 1,
fV (v) = fV,W (v, w) dw1 dw2 (3)
=
1
0
1
w1
4 dw2 dw1 (4)
=
1
0
4(1 − w1) dw1 = 2 (5)
Similarly, for 0 ≤ w1 ≤ w2 ≤ 1,
fW (w) = fV,W (v, w) dv1 dv2 (6)
=
1
0
1
v1
4 dv2 dv1 = 2 (7)
It follows that V and W have PDFs
fV (v) =
2 0 ≤ v1 ≤ v2 ≤ 1
0 otherwise
, fW (w) =
2 0 ≤ w1 ≤ w2 ≤ 1
0 otherwise
(8)
It is easy to verify that fV,W(v, w) = fV(v) fW(w), confirming that V and W are indepen-
dent vectors.
Quiz 5.5
(A) Referring to Theorem 1.19, each test is a subexperiment with three possible out-
comes: L, A and R. In five trials, the vector X = X1 X2 X3 indicating the
number of outcomes of each subexperiment has the multinomial PMF
PX (x) =
⎧
⎨
⎩
5
x1,x2,x3
(0.3)x1(0.6)x2(0.1)x3 x1 + x2 + x3 = 5;
x1, x2, x3 ∈ {0, 1, . . . , 5}
0 otherwise
(1)
We can find the marginal PMF for each Xi from the joint PMF PX(x); however it
is simpler to just start from first principles and observe that X1 is the number of
occurrences of L in five independent tests. If we view each test as a trial with success
probability P[L] = 0.3, we see that X1 is a binomial (n, p) = (5, 0.3) random
variable. Similarly, X2 is a binomial (5, 0.6) random variable and X3 is a binomial
(5, 0.1) random variable. That is, for p1 = 0.3, p2 = 0.6 and p3 = 0.1,
PXi (x) =
5
x px
i (1 − pi )5−x x = 0, 1, . . . , 5
0 otherwise
(2)
35](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-35-320.jpg)
![From the marginal PMFs, we see that X1, X2 and X3 are not independent. Hence, we
must use Theorem 5.6 to find the PMF of W. In particular, since X1 + X2 + X3 = 5
and since each Xi is non-negative, PW (0) = PW (1) = 0. Furthermore,
PW (2) = PX (1, 2, 2) + PX (2, 1, 2) + PX (2, 2, 1) (3)
=
5![0.3(0.6)2(0.1)2 + 0.32(0.6)(0.1)2 + 0.32(0.6)2(0.1)]
2!2!1!
(4)
= 0.1458 (5)
In addition, for w = 3, w = 4, and w = 5, the event W = w occurs if and only if
one of the mutually exclusive events X1 = w, X2 = w, or X3 = w occurs. Thus,
PW (3) = PX1 (3) + PX2 (3) + PX3 (3) = 0.486 (6)
PW (4) = PX1 (4) + PX2 (4) + PX3 (4) = 0.288 (7)
PW (5) = PX1 (5) + PX2 (5) + PX3 (5) = 0.0802 (8)
(B) Since each Yi = 2Xi + 4, we can apply Theorem 5.10 to write
fY (y) =
1
23
fX
y1 − 4
2
,
y2 − 4
2
,
y3 − 4
2
(9)
=
(1/8)e−(y3−4)/2 4 ≤ y1 ≤ y2 ≤ y3
0 otherwise
(10)
Note that for other matrices A, the constraints on y resulting from the constraints
0 ≤ X1 ≤ X2 ≤ X3 can be much more complicated.
Quiz 5.6
We start by finding the components E[Xi ] =
∞
−∞ x fXi (x) dx of µX . To do so, we use
the marginal PDFs fXi (x) found in Quiz 5.3:
E [X1] =
1
0
3x(1 − x)2
dx = 1/4, (1)
E [X2] =
1
0
6x2
(1 − x) dx = 1/2, (2)
E [X3] =
1
0
3x3
dx = 3/4. (3)
To find the correlation matrix RX , we need to find E[Xi X j ] for all i and j. We start with
36](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-36-320.jpg)
![the second moments:
E X2
1 =
1
0
3x2
(1 − x)2
dx = 1/10. (4)
E X2
2 =
1
0
6x3
(1 − x) dx = 3/10. (5)
E X2
3 =
1
0
3x4
dx = 3/5. (6)
Using marginal PDFs from Quiz 5.3, the cross terms are
E [X1 X2] =
∞
−∞
∞
−∞
x1x2 fX1,X2 (x1, x2) , dx1 dx2 (7)
=
1
0
1
x1
6x1x2(1 − x2) dx2 dx1 (8)
=
1
0
[x1 − 3x3
1 + 2x4
1] dx1 = 3/20. (9)
E [X2 X3] =
1
0
1
x2
6x2
2 x3 dx3 dx2 (10)
=
1
0
[3x2
2 − 3x4
2] dx2 = 2/5 (11)
E [X1 X3] =
1
0
1
x1
6x1x3(x3 − x1) dx3 dx1. (12)
=
1
0
(2x1x3
3 − 3x2
1 x2
3)
x3=1
x3=x1
dx1 (13)
=
1
0
[2x1 − 3x2
1 + x4
1] dx1 = 1/5. (14)
Summarizing the results, X has correlation matrix
RX =
⎡
⎣
1/10 3/20 1/5
3/20 3/10 2/5
1/5 2/5 3/5
⎤
⎦ . (15)
Vector X has covariance matrix
CX = RX − E [X] E [X] (16)
=
⎡
⎣
1/10 3/20 1/5
3/20 3/10 2/5
1/5 2/5 3/5
⎤
⎦ −
⎡
⎣
1/4
1/2
3/4
⎤
⎦ 1/4 1/2 3/4 (17)
=
⎡
⎣
1/10 3/20 1/5
3/20 3/10 2/5
1/5 2/5 3/5
⎤
⎦ −
⎡
⎣
1/16 1/8 3/16
1/8 1/4 3/8
3/16 3/8 9/16
⎤
⎦ =
1
80
⎡
⎣
3 2 1
2 4 2
1 2 3
⎤
⎦ . (18)
37](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-37-320.jpg)
![This problem shows that even for fairly simple joint PDFs, computing the covariance matrix
by calculus can be a time consuming task.
Quiz 5.7
We observe that X = AZ + b where
A =
2 1
1 −1
, b =
2
0
. (1)
It follows from Theorem 5.18 that µX = b and that
CX = AA =
2 1
1 −1
2 1
1 −1
=
5 1
1 2
. (2)
Quiz 5.8
First, we observe that Y = AT where A = 1/31 1/31 · · · 1/31 . Since T is a
Gaussian random vector, Theorem 5.16 tells us that Y is a 1 dimensional Gaussian vector,
i.e., just a Gaussian random variable. The expected value of Y is µY = µT = 80. The
covariance matrix of Y is 1 × 1 and is just equal to Var[Y]. Thus, by Theorem 5.16,
Var[Y] = ACT A .
function p=julytemps(T);
[D1 D2]=ndgrid((1:31),(1:31));
CT=36./(1+abs(D1-D2));
A=ones(31,1)/31.0;
CY=(A’)*CT*A;
p=phi((T-80)/sqrt(CY));
In julytemps.m, the first two lines gen-
erate the 31 × 31 covariance matrix CT, or
CT . Next we calculate Var[Y]. The final
step is to use the (·) function to calculate
P[Y < T ].
Here is the output of julytemps.m:
>> julytemps([70 75 80 85 90 95])
ans =
0.0000 0.0221 0.5000 0.9779 1.0000 1.0000
Note that P[T ≤ 70] is not actually zero and that P[T ≤ 90] is not actually 1.0000. Its
just that the MATLAB’s short format output, invoked with the command format short,
rounds off those probabilities. Here is the long format output:
>> format long
>> julytemps([70 75 80 85 90 95])
ans =
Columns 1 through 4
0.00002844263128 0.02207383067604 0.50000000000000 0.97792616932396
Columns 5 through 6
0.99997155736872 0.99999999922010
38](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-38-320.jpg)
![Quiz Solutions – Chapter 6
Quiz 6.1
Let K1, . . . , Kn denote a sequence of iid random variables each with PMF
PK (k) =
1/4 k = 1, . . . , 4
0 otherwise
(1)
We can write Wn in the form of Wn = K1 + · · · + Kn. First, we note that the first two
moments of Ki are
E [Ki ] = (1 + 2 + 3 + 4)/4 = 2.5 (2)
E K2
i = (12
+ 22
+ 32
+ 42
)/4 = 7.5 (3)
Thus the variance of Ki is
Var[Ki ] = E K2
i − (E [Ki ])2
= 7.5 − (2.5)2
= 1.25 (4)
Since E[Ki ] = 2.5, the expected value of Wn is
E [Wn] = E [K1] + · · · + E [Kn] = nE [Ki ] = 2.5n (5)
Since the rolls are independent, the random variables K1, . . . , Kn are independent. Hence,
by Theorem 6.3, the variance of the sum equals the sum of the variances. That is,
Var[Wn] = Var[K1] + · · · + Var[Kn] = 1.25n (6)
Quiz 6.2
Random variables X and Y have PDFs
fX (x) =
3e−3x x ≥ 0
0 otherwise
fY (y) =
2e−2y y ≥ 0
0 otherwise
(1)
Since X and Y are nonnegative, W = X + Y is nonnegative. By Theorem 6.5, the PDF of
W = X + Y is
fW (w) =
∞
−∞
fX (w − y) fY (y) dy = 6
w
0
e−3(w−y)
e−2y
dy (2)
Fortunately, this integral is easy to evaluate. For w > 0,
fW (w) = e−3w
ey w
0
= 6 e−2w
− e−3w
(3)
Since fW (w) = 0 for w < 0, a conmplete expression for the PDF of W is
fW (w) =
6e−2w 1 − e−w w ≥ 0,
0 otherwise.
(4)
40](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-40-320.jpg)
![Quiz 6.3
The MGF of K is
φK (s) = E esK
==
4
k=0
(0.2)esk
= 0.2 1 + es
+ e2s
+ e3s
+ e4s
(1)
We find the moments by taking derivatives. The first derivative of φK (s) is
dφK (s)
ds
= 0.2(es
+ 2e2s
+ 3e3s
+ 4e4s
) (2)
Evaluating the derivative at s = 0 yields
E [K] =
dφK (s)
ds s=0
= 0.2(1 + 2 + 3 + 4) = 2 (3)
To find higher-order moments, we continue to take derivatives:
E K2
=
d2φK (s)
ds2
s=0
= 0.2(es
+ 4e2s
+ 9e3s
+ 16e4s
)
s=0
= 6 (4)
E K3
=
d3φK (s)
ds3
s=0
= 0.2(es
+ 8e2s
+ 27e3s
+ 64e4s
)
s=0
= 20 (5)
E K4
=
d4φK (s)
ds4
s=0
= 0.2(es
+ 16e2s
+ 81e3s
+ 256e4s
)
s=0
= 70.8 (6)
(7)
Quiz 6.4
(A) Each Ki has MGF
φK (s) = E esKi
=
es + e2s + · · · + ens
n
=
es(1 − ens)
n(1 − es)
(1)
Since the sequence of Ki is independent, Theorem 6.8 says the MGF of J is
φJ (s) = (φK (s))m
=
ems(1 − ens)m
nm(1 − es)m
(2)
(B) Since the set of α j X j are independent Gaussian random variables, Theorem 6.10
says that W is a Gaussian random variable. Thus to find the PDF of W, we need
only find the expected value and variance. Since the expectation of the sum equals
the sum of the expectations:
E [W] = αE [X1] + α2
E [X2] + · · · + αn
E [Xn] = 0 (3)
41](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-41-320.jpg)
![Since the α j X j are independent, the variance of the sum equals the sum of the vari-
ances:
Var[W] = α2
Var[X1] + α4
Var[X2] + · · · + α2n
Var[Xn] (4)
= α2
+ 2(α2
)2
+ 3(α2
)3
+ · · · + n(α2
)n
(5)
Defining q = α2, we can use Math Fact B.6 to write
Var[W] =
α2 − α2n+2[1 + n(1 − α2)]
(1 − α2)2
(6)
With E[W] = 0 and σ2
W = Var[W], we can write the PDF of W as
fW (w) =
1
2πσ2
W
e−w2/2σ2
W (7)
Quiz 6.5
(1) From Table 6.1, each Xi has MGF φX (s) and random variable N has MGF φN (s)
where
φX (s) =
1
1 − s
, φN (s) =
1
5es
1 − 4
5es
. (1)
From Theorem 6.12, R has MGF
φR(s) = φN (ln φX (s)) =
1
5φX (s)
1 − 4
5φX (s)
(2)
Substituting the expression for φX (s) yields
φR(s) =
1
5
1
5 − s
. (3)
(2) From Table 6.1, we see that R has the MGF of an exponential (1/5) random variable.
The corresponding PDF is
fR (r) =
(1/5)e−r/5 r ≥ 0
0 otherwise
(4)
This quiz is an example of the general result that a geometric sum of exponential
random variables is an exponential random variable.
42](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-42-320.jpg)
![Quiz 6.6
(1) The expected access time is
E [X] =
∞
−∞
x fX (x) dx =
12
0
x
12
dx = 6 msec (1)
(2) The second moment of the access time is
E X2
=
∞
−∞
x2
fX (x) dx =
12
0
x2
12
dx = 48 (2)
The variance of the access time is Var[X] = E[X2] − (E[X])2 = 48 − 36 = 12.
(3) Using Xi to denote the access time of block i, we can write
A = X1 + X2 + · · · + X12 (3)
Since the expectation of the sum equals the sum of the expectations,
E [A] = E [X1] + · · · + E [X12] = 12E [X] = 72 msec (4)
(4) Since the Xi are independent,
Var[A] = Var[X1] + · · · + Var[X12] = 12 Var[X] = 144 (5)
Hence, the standard deviation of A is σA = 12
(5) To use the central limit theorem, we write
P [A > 75] = 1 − P [A ≤ 75] (6)
= 1 − P
A − E [A]
σA
≤
75 − E [A]
σA
(7)
≈ 1 −
75 − 72
12
(8)
= 1 − 0.5987 = 0.4013 (9)
Note that we used Table 3.1 to look up (0.25).
(6) Once again, we use the central limit theorem and Table 3.1 to estimate
P [A < 48] = P
A − E [A]
σA
<
48 − E [A]
σA
(10)
≈
48 − 72
12
(11)
= 1 − (2) = 1 − 0.9773 = 0.0227 (12)
43](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-43-320.jpg)
![Quiz 6.7
Random variable Kn has a binomial distribution for n trials and success probability
P[V ] = 3/4.
(1) The expected number of voice calls out of 48 calls is E[K48] = 48P[V ] = 36.
(2) The variance of K48 is
Var[K48] = 48P [V ] (1 − P [V ]) = 48(3/4)(1/4) = 9 (1)
Thus K48 has standard deviation σK48 = 3.
(3) Using the ordinary central limit theorem and Table 3.1 yields
P [30 ≤ K48 ≤ 42] ≈
42 − 36
3
−
30 − 36
3
= (2) − (−2) (2)
Recalling that (−x) = 1 − (x), we have
P [30 ≤ K48 ≤ 42] ≈ 2 (2) − 1 = 0.9545 (3)
(4) Since K48 is a discrete random variable, we can use the De Moivre-Laplace approx-
imation to estimate
P [30 ≤ K48 ≤ 42] ≈
42 + 0.5 − 36
3
−
30 − 0.5 − 36
3
(4)
= 2 (2.16666) − 1 = 0.9687 (5)
Quiz 6.8
The train interarrival times X1, X2, X3 are iid exponential (λ) random variables. The
arrival time of the third train is
W = X1 + X2 + X3. (1)
In Theorem 6.11, we found that the sum of three iid exponential (λ) random variables is an
Erlang (n = 3, λ) random variable. From Appendix A, we find that W has expected value
and variance
E [W] = 3/λ = 6 Var[W] = 3/λ2
= 12 (2)
(1) By the Central Limit Theorem,
P [W > 20] = P
W − 6
√
12
>
20 − 6
√
12
≈ Q(7/
√
3) = 2.66 × 10−5
(3)
44](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-44-320.jpg)
![(2) To use the Chernoff bound, we note that the MGF of W is
φW (s) =
λ
λ − s
3
=
1
(1 − 2s)3
(4)
The Chernoff bound states that
P [W > 20] ≤ min
s≥0
e−20s
φX (s) = min
s≥0
e−20s
(1 − 2s)3
(5)
To minimize h(s) = e−20s/(1 − 2s)3, we set the derivative of h(s) to zero:
dh(s)
ds
=
−20(1 − 2s)3e−20s + 6e−20s(1 − 2s)2
(1 − 2s)6
= 0 (6)
This implies 20(1 − 2s) = 6 or s = 7/20. Applying s = 7/20 into the Chernoff
bound yields
P [W > 20] ≤
e−20s
(1 − 2s)3
s=7/20
= (10/3)3
e−7
= 0.0338 (7)
(3) Theorem 3.11 says that for any w > 0, the CDF of the Erlang (λ, 3) random variable
W satisfies
FW (w) = 1 −
2
k=0
(λw)ke−λw
k!
(8)
Equivalently, for λ = 1/2 and w = 20,
P [W > 20] = 1 − FW (20) (9)
= e−10
1 +
10
1!
+
102
2!
= 61e−10
= 0.0028 (10)
Although the Chernoff bound is relatively weak in that it overestimates the proba-
bility by roughly a factor of 12, it is a valid bound. By contrast, the Central Limit
Theorem approximation grossly underestimates the true probability.
Quiz 6.9
One solution to this problem is to follow the approach of Example 6.19:
%unifbinom100.m
sx=0:100;sy=0:100;
px=binomialpmf(100,0.5,sx); py=duniformpmf(0,100,sy);
[SX,SY]=ndgrid(sx,sy); [PX,PY]=ndgrid(px,py);
SW=SX+SY; PW=PX.*PY;
sw=unique(SW); pw=finitepmf(SW,PW,sw);
pmfplot(sw,pw,’itw’,’itP_W(w)’);
A graph of the PMF PW (w) appears in Figure 2 With some thought, it should be apparent
that the finitepmf function is implementing the convolution of the two PMFs.
45](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-45-320.jpg)
![Quiz Solutions – Chapter 7
Quiz 7.1
An exponential random variable with expected value 1 also has variance 1. By Theo-
rem 7.1, Mn(X) has variance Var[Mn(X)] = 1/n. Hence, we need n = 100 samples.
Quiz 7.2
The arrival time of the third elevator is W = X1 + X2 + X3. Since each Xi is uniform
(0, 30),
E [Xi ] = 15, Var [Xi ] =
(30 − 0)2
12
= 75. (1)
Thus E[W] = 3E[Xi ] = 45, and Var[W] = 3 Var[Xi ] = 225.
(1) By the Markov inequality,
P [W > 75] ≤
E [W]
75
=
45
75
=
3
5
(2)
(2) By the Chebyshev inequality,
P [W > 75] = P [W − E [W] > 30] (3)
≤ P [|W − E [W]| > 30] ≤
Var [W]
302
=
225
900
=
1
4
(4)
Quiz 7.3
Define the random variable W = (X − µX )2. Observe that V100(X) = M100(W). By
Theorem 7.6, the mean square error is
E (M100(W) − µW )2
=
Var[W]
100
(1)
Observe that µX = 0 so that W = X2. Thus,
µW = E X2
=
1
−1
x2
fX (x) dx = 1/3 (2)
E W2
= E X4
=
1
−1
x4
fX (x) dx = 1/5 (3)
Therefore Var[W] = E[W2] − µ2
W = 1/5 − (1/3)2 = 4/45 and the mean square error is
4/4500 = 0.000889.
47](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-47-320.jpg)
![Quiz 7.4
Assuming the number n of samples is large, we can use a Gaussian approximation for
Mn(X). SinceE[X] = p and Var[X] = p(1 − p), we apply Theorem 7.13 which says that
the interval estimate
Mn(X) − c ≤ p ≤ Mn(X) + c (1)
has confidence coefficient 1 − α where
α = 2 − 2
c
√
n
p(1 − p)
. (2)
We must ensure for every value of p that 1 − α ≥ 0.9 or α ≤ 0.1. Equivalently, we must
have
c
√
n
p(1 − p)
≥ 0.95 (3)
for every value of p. Since (x) is an increasing function of x, we must satisfy c
√
n ≥
1.65p(1 − p). Since p(1 − p) ≤ 1/4 for all p, we require that
c ≥
1.65
4
√
n
=
0.41
√
n
. (4)
The 0.9 confidence interval estimate of p is
Mn(X) −
0.41
√
n
≤ p ≤ Mn(X) +
0.41
√
n
. (5)
For the 0.99 confidence interval, we have α ≤ 0.01, implying (c
√
n/(p(1−p))) ≥ 0.995.
This implies c
√
n ≥ 2.58p(1 − p). Since p(1 − p) ≤ 1/4 for all p, we require that
c ≥ (0.25)(2.58)/
√
n. In this case, the 0.99 confidence interval estimate is
Mn(X) −
0.645
√
n
≤ p ≤ Mn(X) +
0.645
√
n
. (6)
Note that if M100(X) = 0.4, then the 0.99 confidence interval estimate is
0.3355 ≤ p ≤ 0.4645. (7)
The interval is wide because the 0.99 confidence is high.
Quiz 7.5
Following the approach of bernoullitraces.m, we generate m = 1000 sample
paths, each sample path having n = 100 Bernoulli traces. at time k, OK(k) counts the
fraction of sample paths that have sample mean within one standard error of p. The pro-
gram bernoullisample.m generates graphs the number of traces within one standard
error as a function of the time, i.e. the number of trials in each trace.
48](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-48-320.jpg)
![Quiz Solutions – Chapter 8
Quiz 8.1
From the problem statement, each Xi has PDF and CDF
fXi (x) =
e−x x ≥ 0
0 otherwise
FXi (x) =
0 x < 0
1 − e−x x ≥ 0
(1)
Hence, the CDF of the maximum of X1, . . . , X15 obeys
FX (x) = P [X ≤ x] = P [X1 ≤ x, X2 ≤ x, · · · , X15 ≤ x] = [P [Xi ≤ x]]15
. (2)
This implies that for x ≥ 0,
FX (x) = FXi (x)
15
= 1 − e−x 15
(3)
To design a significance test, we must choose a rejection region for X. A reasonable choice
is to reject the hypothesis if X is too small. That is, let R = {X ≤ r}. For a significance
level of α = 0.01, we obtain
α = P [X ≤ r] = (1 − e−r
)15
= 0.01 (4)
It is straightforward to show that
r = − ln 1 − (0.01)1/15
= 1.33 (5)
Hence, if we observe X < 1.33, then we reject the hypothesis.
Quiz 8.2
From the problem statement, the conditional PMFs of K are
PK|H0 (k) =
104ke−104
k! k = 0, 1, . . .
0 otherwise
(1)
PK|H1 (k) =
106ke−106
k! k = 0, 1, . . .
0 otherwise
(2)
Since the two hypotheses are equally likely, the MAP and ML tests are the same. From
Theorem 8.6, the ML hypothesis rule is
k ∈ A0 if PK|H0 (k) ≥ PK|H1 (k) ; k ∈ A1 otherwise. (3)
This rule simplifies to
k ∈ A0 if k ≤ k∗
=
106 − 104
ln 100
= 214, 975.7; k ∈ A1 otherwise. (4)
Thus if we observe at least 214, 976 photons, then we accept hypothesis H1.
50](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-50-320.jpg)
![Quiz 8.3
For the QPSK system, a symbol error occurs when si is transmitted but (X1, X2) ∈ Aj
for some j = i. For a QPSK system, it is easier to calculate the probability of a correct
decision. Given H0, the conditional probability of a correct decision is
P [C|H0] = P [X1 > 0, X2 > 0|H0] = P
√
E/2 + N1 > 0,
√
E/2 + N2 > 0 (1)
Because of the symmetry of the signals, P[C|H0] = P[C|Hi ] for all i. This implies the
probability of a correct decision is P[C] = P[C|H0]. Since N1 and N2 are iid Gaussian
(0, σ) random variables, we have
P [C] = P [C|H0] = P
√
E/2 + N1 > 0 P
√
E/2 + N2 > 0 (2)
= P N1 > −
√
E/2
2
(3)
= 1 −
−
√
E/2
σ
2
(4)
Since (−x) = 1 − (x), we have P[C] = 2( E/2σ2). Equivalently, the probability
of error is
PERR = 1 − P [C] = 1 − 2 E
2σ2
(5)
Quiz 8.4
To generate the ROC, the existing program sqdistor already calculates this miss
probability PMISS = P01 and the false alarm probability PFA = P10. The modified pro-
gram, sqdistroc.m is essentially the same as sqdistor except the output is a ma-
trix FM whose columns are the false alarm and miss probabilities. Next, the program
sqdistrocplot.m calls sqdistroc three times to generate a plot that compares the
receiver performance for the three requested values of d. Here is the modified code:
function FM=sqdistroc(v,d,m,T)
%square law distortion recvr
%P(error) for m bits tested
%transmit v volts or -v volts,
%add N volts, N is Gauss(0,1)
%add d(v+N)ˆ2 distortion
%receive 1 if x>T, otherwise 0
%FM = [P(FA) P(MISS)]
x=(v+randn(m,1));
[XX,TT]=ndgrid(x,T(:));
P01=sum((XX+d*(XX.ˆ2)< TT),1)/m;
x= -v+randn(m,1);
[XX,TT]=ndgrid(x,T(:));
P10=sum((XX+d*(XX.ˆ2)>TT),1)/m;
FM=[P10(:) P01(:)];
function FM=sqdistrocplot(v,m,T);
FM1=sqdistroc(v,0.1,m,T);
FM2=sqdistroc(v,0.2,m,T);
FM5=sqdistroc(v,0.3,m,T);
FM=[FM1 FM2 FM5];
loglog(FM1(:,1),FM1(:,2),’-k’, ...
FM2(:,1),FM2(:,2),’--k’, ...
FM5(:,1),FM5(:,2),’:k’);
legend(’it d=0.1’,’it d=0.2’,...
’it d=0.3’,3)
ylabel(’P_{MISS}’);
xlabel(’P_{FA}’);
51](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-51-320.jpg)
![Quiz Solutions – Chapter 9
Quiz 9.1
(1) First, we calculate the marginal PDF for 0 ≤ y ≤ 1:
fY (y) =
y
0
2(y + x) dx = 2xy + x2
x=y
x=0
= 3y2
(1)
This implies the conditional PDF of X given Y is
fX|Y (x|y) =
fX,Y (x, y)
fY (y)
=
2
3y + 2x
3y2 0 ≤ x ≤ y
0 otherwise
(2)
(2) The minimum mean square error estimate of X given Y = y is
ˆxM(y) = E [X|Y = y] =
y
0
2x
3y
+
2x2
3y2
dx = 5y/9 (3)
Thus the MMSE estimator of X given Y is ˆXM(Y) = 5Y/9.
(3) To obtain the conditional PDF fY|X (y|x), we need the marginal PDF fX (x). For
0 ≤ x ≤ 1,
fX (x) =
1
x
2(y + x) dy = y2
+ 2xy
y=1
y=x
= 1 + 2x − 3x2
(4)
(5)
For 0 ≤ x ≤ 1, the conditional PDF of Y given X is
fY|X (y|x) =
2(y+x)
1+2x−3x2 x ≤ y ≤ 1
0 otherwise
(6)
(4) The MMSE estimate of Y given X = x is
ˆyM(x) = E [Y|X = x] =
1
x
2y2 + 2xy
1 + 2x − 3x2
dy (7)
=
2y3/3 + xy2
1 + 2x − 3x2
y=1
y=x
(8)
=
2 + 3x − 5x3
3 + 6x − 9x2
(9)
53](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-53-320.jpg)
![Quiz 9.2
(1) Since the expectation of the sum equals the sum of the expectations,
E [R] = E [T ] + E [X] = 0 (1)
(2) Since T and X are independent, the variance of the sum R = T + X is
Var[R] = Var[T ] + Var[X] = 9 + 3 = 12 (2)
(3) Since T and R have expected values E[R] = E[T ] = 0,
Cov [T, R] = E [T R] = E [T (T + X)] = E T 2
+ E [T X] (3)
Since T and X are independent and have zero expected value, E[T X] = E[T ]E[X] =
0 and E[T 2] = Var[T ]. Thus Cov[T, R] = Var[T ] = 9.
(4) From Definition 4.8, the correlation coefficient of T and R is
ρT,R =
Cov [T, R]
√
Var[R] Var[T ]
=
σT
σR
=
√
3/2 (4)
(5) From Theorem 9.4, the optimum linear estimate of T given R is
ˆTL(R) = ρT,R
σT
σR
(R − E [R]) + E [T ] (5)
Since E[R] = E[T ] = 0 and ρT,R = σT /σR,
ˆTL(R) =
σ2
T
σ2
R
R =
σ2
T
σ2
T + σ2
X
R =
3
4
R (6)
Hence a∗ = 3/4 and b∗ = 0.
(6) By Theorem 9.4, the mean square error of the linear estimate is
e∗
L = Var[T ](1 − ρ2
T,R) = 9(1 − 3/4) = 9/4 (7)
Quiz 9.3
When R = r, the conditional PDF of X = Y −40−40 log10 r is Gaussian with expected
value −40 − 40 log10 r and variance 64. The conditional PDF of X given R is
fX|R (x|r) =
1
√
128π
e−(x+40+40 log10 r)2/128
(1)
54](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-54-320.jpg)
![Quiz 9.4
(1) From Theorem 9.4, the LMSE estimate of X2 given Y2 is ˆX2(Y2) = a∗Y2 +b∗ where
a∗
=
Cov [X2, Y2]
Var[Y2]
, b∗
= µX2 − a∗
µY2. (1)
Because E[X] = E[Y] = 0,
Cov [X2, Y2] = E [X2Y2] = E [X2(X2 + W2)] = E X2
2 = 1 (2)
Var[Y2] = Var[X2] + Var[W2] = E X2
2 + E W2
2 = 1.1 (3)
It follows that a∗ = 1/1.1. Because µX2 = µY2 = 0, it follows that b∗ = 0. Finally,
to compute the expected square error, we calculate the correlation coefficient
ρX2,Y2 =
Cov [X2, Y2]
σX2σY2
=
1
√
1.1
(4)
The expected square error is
e∗
L = Var[X2](1 − ρ2
X2,Y2
) = 1 −
1
1.1
=
1
11
= 0.0909 (5)
(2) Since Y = X + W and E[X] = E[W] = 0, it follows that E[Y] = 0. Thus we can
apply Theorem 9.7. Note that X and W have correlation matrices
RX =
1 −0.9
−0.9 1
, RW =
0.1 0
0 0.1
. (6)
In terms of Theorem 9.7, n = 2 and we wish to estimate X2 given the observation
vector Y = Y1 Y2 . To apply Theorem 9.7, we need to find RY and RYX2.
RY = E YY = E (X + W)(X + W ) (7)
= E XX + XW + WX + WW . (8)
Because X and W are independent, E[XW ] = E[X]E[W ] = 0. Similarly, E[WX ] =
0. This implies
RY = E XX + E WW = RX + RW =
1.1 −0.9
−0.9 1.1
. (9)
In addition, we need to find
RYX2 = E [YX2] =
E [Y1 X2]
E [Y2 X2]
=
E [(X1 + W1)X2]
E [(X2 + W2)X2]
. (10)
56](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-56-320.jpg)
![Since X and W are independent vectors, E[W1 X2] = E[W1]E[X2] = 0 and E[W2 X2] =
0. Thus
RYX2 =
E[X1 X2]
E X2
2
=
−0.9
1
. (11)
By Theorem 9.7,
ˆa = R−1
Y RYX2 =
−0.225
0.725
(12)
Therefore, the optimum linear estimator of X2 given Y1 and Y2 is
ˆXL = ˆa Y = −0.225Y1 + 0.725Y2. (13)
The mean square error is
Var [X2] − ˆa RYX2 = Var [X] − a1rY1,X2 − a2rY2,X2 = 0.0725. (14)
Quiz 9.5
Since X and W have zero expected value, Y also has zero expected value. Thus, by
Theorem 9.7, ˆXL(Y) = ˆa Y where ˆa = R−1
Y RYX . Since X and W are independent,
E[WX] = 0 and E[XW ] = 0 . This implies
RYX = E [YX] = E [(1X + W)X] = 1E X2
= 1. (1)
By the same reasoning, the correlation matrix of Y is
RY = E YY = E (1X + W)(1 X + W ) (2)
= 11 E X2
+ 1E XW + E [WX] 1 + E WW (3)
= 11 + RW (4)
Note that 11 is a 20 × 20 matrix with every entry equal to 1. Thus,
ˆa = R−1
Y RYX = 11 + RW
−1
1 (5)
and the optimal linear estimator is
ˆXL(Y) = 1 11 + RW
−1
Y (6)
The mean square error is
e∗
L = Var[X] − ˆa RYX = 1 − 1 11 + RW
−1
1 (7)
Now we note that RW has i, jth entry RW(i, j) = c|i− j|−1. The question we must address
is what value c minimizes e∗
L. This problem is atypical in that one does not usually get
57](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-57-320.jpg)
![to choose the correlation structure of the noise. However, we will see that the answer is
somewhat instructive.
We note that the answer is not obviously apparent from Equation (7). In particular, we
observe that Var[Wi ] = RW(i, i) = 1/c. Thus, when c is small, the noises Wi have high
variance and we would expect our estimator to be poor. On the other hand, if c is large
Wi and Wj are highly correlated and the separate measurements of X are very dependent.
This would suggest that large values of c will also result in poor MSE. If this argument is
not clear, consider the extreme case in which every Wi and Wj have correlation coefficient
ρi j = 1. In this case, our 20 measurements will be all the same and one measurement is as
good as 20 measurements.
To find the optimal value of c, we write a MATLAB function mquiz9(c) to calculate
the MSE for a given c and second function that finds plots the MSE for a range of values
of c.
function [mse,af]=mquiz9(c);
v1=ones(20,1);
RW=toeplitz(c.ˆ((0:19)-1));
RY=(v1*(v1’)) +RW;
af=(inv(RY))*v1;
mse=1-((v1’)*af);
function cmin=mquiz9minc(c);
msec=zeros(size(c));
for k=1:length(c),
[msec(k),af]=mquiz9(c(k));
end
plot(c,msec);
xlabel(’c’);ylabel(’e_Lˆ*’);
[msemin,optk]=min(msec);
cmin=c(optk);
Note in mquiz9 that v1 corresponds to the vector 1 of all ones. The following commands
finds the minimum c and also produces the following graph:
>> c=0.01:0.01:0.99;
>> mquiz9minc(c)
ans =
0.4500
0 0.5 1
0.2
0.4
0.6
0.8
1
c
e
L
*
As we see in the graph, both small values and large values of c result in large MSE.
58](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-58-320.jpg)
![Quiz Solutions – Chapter 10
Quiz 10.1
There are many correct answers to this question. A correct answer specifies enough
random variables to specify the sample path exactly. One choice for an alternate set of
random variables that would specify m(t, s) is
• m(0, s), the number of ongoing calls at the start of the experiment
• N, the number of new calls that arrive during the experiment
• X1, . . . , XN , the interarrival times of the N new arrivals
• H, the number of calls that hang up during the experiment
• D1, . . . , DH , the call completion times of the H calls that hang up
Quiz 10.2
(1) We obtain a continuous time, continuous valued process when we record the temper-
ature as a continuous waveform over time.
(2) If at every moment in time, we round the temperature to the nearest degree, then we
obtain a continuous time, discrete valued process.
(3) If we sample the process in part (a) every T seconds, then we obtain a discrete time,
continuous valued process.
(4) Rounding the samples in part (c) to the nearest integer degree yields a discrete time,
discrete valued process.
Quiz 10.3
(1) Each resistor has resistance R in ohms with uniform PDF
fR (r) =
0.01 950 ≤ r ≤ 1050
0 otherwise
(1)
The probability that a test produces a 1% resistor is
p = P [990 ≤ R ≤ 1010] =
1010
990
(0.01) dr = 0.2 (2)
59](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-59-320.jpg)
![(2) In t seconds, exactly t resistors are tested. Each resistor is a 1% resistor with proba-
bility p, independent of any other resistor. Consequently, the number of 1% resistors
found has the binomial PMF
PN(t) (n) =
t
n pn(1 − p)t−n n = 0, 1, . . . , t
0 otherwise
(3)
(3) First we will find the PMF of T1. This problem is easy if we view each resistor test
as an independent trial. A success occurs on a trial with probability p if we find a
1% resistor. The first 1% resistor is found at time T1 = t if we observe failures on
trials 1, . . . , t − 1 followed by a success on trial t. Hence, just as in Example 2.11,
T1 has the geometric PMF
PT1 (t) =
(1 − p)t−1 p t = 1, 2, . . .
9 otherwise
(4)
Since p = 0.2, the probability the first 1% resistor is found in exactly five seconds is
PT1(5) = (0.8)4(0.2) = 0.08192.
(4) From Theorem 2.5, a geometric random variable with success probability p has ex-
pected value 1/p. In this problem, E[T1] = 1/p = 5.
(5) Note that once we find the first 1% resistor, the number of additional trials needed to
find the second 1% resistor once again has a geometric PMF with expected value 1/p
since each independent trial is a success with probability p. That is, T2 = T1 + T
where T is independent and identically distributed to T1. Thus
E [T2|T1 = 10] = E [T1|T1 = 10] + E T |T1 = 10 (5)
= 10 + E T = 10 + 5 = 15 (6)
Quiz 10.4
Since each Xi is a N(0, 1) random variable, each Xi has PDF
fX(i) (x) =
1
√
2π
e−x2/2
(1)
By Theorem 10.1, the joint PDF of X = X1 · · · Xn is
fX (x) = fX(1),...,X(n) (x1, . . . , xn) =
k
i=1
fX (xi ) =
1
(2π)n/2
e−(x2
1+···+x2
n)/2
(2)
60](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-60-320.jpg)
![Quiz 10.5
The first and second hours are nonoverlapping intervals. Since one hour equals 3600
sec and the Poisson process has a rate of 10 packets/sec, the expected number of packets
in each hour is E[Mi ] = α = 36, 000. This implies M1 and M2 are independent Poisson
random variables each with PMF
PMi (m) =
αme−α
m! m = 0, 1, 2, . . .
0 otherwise
(1)
Since M1 and M2 are independent, the joint PMF of M1 and M2 is
PM1,M2 (m1, m2) = PM1 (m1) PM2 (m2) =
⎧
⎪⎪⎨
⎪⎪⎩
αm1+m2e−2α
m1!m2! m1 = 0, 1, . . . ;
m2 = 0, 1, . . . ,
0 otherwise.
(2)
Quiz 10.6
To answer whether N (t) is a Poisson process, we look at the interarrival times. Let
X1, X2, . . . denote the interarrival times of the N(t) process. Since we count only even-
numbered arrival for N (t), the time until the first arrival of the N (t) is Y1 = X1 + X2.
Since X1 and X2 are independent exponential (λ) random variables, Y1 is an Erlang (n =
2, λ) random variable; see Theorem 6.11. Since Yi (t), the ith interarrival time of the N (t)
process, has the same PDF as Y1(t), we can conclude that the interarrival times of N (t)
are not exponential random variables. Thus N (t) is not a Poisson process.
Quiz 10.7
First, we note that for t > s,
X(t) − X(s) =
W(t) − W(s)
√
α
(1)
Since W(t) − W(s) is a Gaussian random variable, Theorem 3.13 states that W(t) − W(s)
is Gaussian with expected value
E [X(t) − X(s)] =
E [W(t) − W(s)]
√
α
= 0 (2)
and variance
E (W(t) − W(s))2
=
E (W(t) − W(s))2
α
=
α(t − s)
α
(3)
Consider s ≤ s < t. Since s ≥ s , W(t) − W(s) is independent of W(s ). This implies
[W(t) − W(s)]/
√
α is independent of W(s )/
√
α for all s ≥ s . That is, X(t) − X(s) is
independent of X(s ) for all s ≥ s . Thus X(t) is a Brownian motion process with variance
Var[X(t)] = t.
61](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-61-320.jpg)
![Quiz 10.8
First we find the expected value
µY (t) = µX (t) + µN (t) = µX (t). (1)
To find the autocorrelation, we observe that since X(t) and N(t) are independent and since
N(t) has zero expected value, E[X(t)N(t )] = E[X(t)]E[N(t )] = 0. Since RY (t, τ) =
E[Y(t)Y(t + τ)], we have
RY (t, τ) = E [(X(t) + N(t)) (X(t + τ) + N(t + τ))] (2)
= E [X(t)X(t + τ)] + E [X(t)N(t + τ)]
+ E [X(t + τ)N(t)] + E [N(t)N(t + τ)] (3)
= RX (t, τ) + RN (t, τ). (4)
Quiz 10.9
From Definition 10.14, X1, X2, . . . is a stationary random sequence if for all sets of
time instants n1, . . . , nm and time offset k,
fXn1
,...,Xnm
(x1, . . . , xm) = fXn1+k,...,Xnm+k (x1, . . . , xm) (1)
Since the random sequence is iid,
fXn1
,...,Xnm
(x1, . . . , xm) = fX (x1) fX (x2) · · · fX (xm) (2)
Similarly, for time instants n1 + k, . . . , nm + k,
fXn1+k,...,Xnm+k (x1, . . . , xm) = fX (x1) fX (x2) · · · fX (xm) (3)
We can conclude that the iid random sequence is stationary.
Quiz 10.10
We must check whether each function R(τ) meets the conditions of Theorem 10.12:
R(τ) ≥ 0 R(τ) = R(−τ) |R(τ)| ≤ R(0) (1)
(1) R1(τ) = e−|τ| meets all three conditions and thus is valid.
(2) R2(τ) = e−τ2
also is valid.
(3) R3(τ) = e−τ cos τ is not valid because
R3(−2π) = e2π
cos 2π = e2π
> 1 = R3(0) (2)
(4) R4(τ) = e−τ2
sin τ also cannot be an autocorrelation function because
R4(π/2) = e−π/2
sin π/2 = e−π/2
> 0 = R4(0) (3)
62](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-62-320.jpg)
![Quiz 10.11
(1) The autocorrelation of Y(t) is
RY (t, τ) = E [Y(t)Y(t + τ)] (1)
= E [X(−t)X(−t − τ)] (2)
= RX (−t − (−t − τ)) = RX (τ) (3)
Since E[Y(t)] = E[X(−t)] = µX , we can conclude that Y(t) is a wide sense
stationary process. In fact, we see that by viewing a process backwards in time, we
see the same second order statistics.
(2) Since X(t) and Y(t) are both wide sense stationary processes, we can check whether
they are jointly wide sense stationary by seeing if RXY (t, τ) is just a function of τ.
In this case,
RXY (t, τ) = E [X(t)Y(t + τ)] (4)
= E [X(t)X(−t − τ)] (5)
= RX (t − (−t − τ)) = RX (2t + τ) (6)
Since RXY (t, τ) depends on both t and τ, we conclude that X(t) and Y(t) are not
jointly wide sense stationary. To see why this is, suppose RX (τ) = e−|τ| so that
samples of X(t) far apart in time have almost no correlation. In this case, as t gets
larger, Y(t) = X(−t) and X(t) become less and less correlated.
Quiz 10.12
From the problem statement,
E [X(t)] = E [X(t + 1)] = 0 (1)
E [X(t)X(t + 1)] = 1/2 (2)
Var[X(t)] = Var[X(t + 1)] = 1 (3)
The Gaussian random vector X = X(t) X(t + 1) has covariance matrix and corre-
sponding inverse
CX =
1 1/2
1/2 1
C−1
X =
4
3
1 −1/2
−1/2 1
(4)
Since
x C−1
X x = x0 x1
4
3
1 −1/2
−1/2 1
x0
x1
=
4
3
x2
0 − x0x+x2
1 (5)
the joint PDF of X(t) and X(t + 1) is the Gaussian vector PDF
fX(t),X(t+1) (x0, x1) =
1
(2π)n/2[det (CX)]1/2
exp −
1
2
x C−1
X x (6)
=
1
√
3π2
e−2
3 x2
0−x0x1+x2
1 (7)
63](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-63-320.jpg)
![Thus we know that M(t) will stay the same until then. In our simulation, we use the vector
t as the set of time instances at which we inspect the system state. Thus for all times t(i)
between the current head-of-schedule event and the next, we set m(i) to the current switch
state.
The complete program is shown in Figure 5. In most programming languages, it is
common to implement the event schedule as a linked list where each item in the list has
a data structure indicating an event timestamp and the type of the event. In MATLAB, a
simple (but not elegant) way to do this is to have maintain two vectors: time is a list
of timestamps of scheduled events and event is a the list of event types. In this case,
event(i)=1 if the ith scheduled event is an arrival, or event(i)=-1 if the ith sched-
uled event is a departure.
When the program is passed a vector t, the output [m a b] is such that m(i) is the
number of ongoing calls at time t(i) while a and b are the number of admits and blocks.
The following instructions
t=0:0.1:5000;
[m,a,b]=simblockswitch(10,0.1,120,t);
plot(t,m);
generated a simulation lasting 5,000 minutes. A sample path of the first 100 minutes of
that simulation is shown in Figure 4. The 5,000 minute full simulation produced a=49658
admitted calls and b=239 blocked calls. We can estimate the probability a call is blocked
as
ˆPb =
b
a + b
= 0.0048. (1)
In Chapter 12, we will learn that the exact blocking probability is given by Equation (12.93),
a result known as the “Erlang-B formula.” From the Erlang-B formula, we can calculate
that the exact blocking probability is Pb = 0.0057. One reason our simulation underesti-
mates the blocking probability is that in a 5,000 minute simulation, roughly the first 100
minutes are needed to load up the switch since the switch is idle when the simulation starts
at time t = 0. However, this says that roughly the first two percent of the simulation time
was unusual. Thus this would account for only part of the disparity. The rest of the gap
between 0.0048 and 0.0057 is that a simulation that includes only 239 blocks is not all that
likely to give a very accurate result for the blocking probability.
Note that in Chapter 12, we will learn that the blocking switch is an example of an
M/M/c/c queue, a kind of Markov chain. Chapter 12 develops techniques for analyzing
and simulating systems described by Markov chains that are much simpler than the discrete
event simulation technique shown here. Nevertheless, for very complicated systems, the
discrete event simulation is widely-used and often very efficient simulation method.
65](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-65-320.jpg)
![function [M,admits,blocks]=simblockswitch(lam,mu,c,t);
blocks=0; %total # blocks
admits=0; %total # admits
M=zeros(size(t));
n=0; % # in system
time=[ exponentialrv(lam,1) ];
event=[ 1 ]; %first event is an arrival
timenow=0;
tmax=max(t);
while (timenow<tmax)
M((timenow<=t)&(t<time(1)))=n;
timenow=time(1);
eventnow=event(1);
event(1)=[ ]; time(1)= [ ]; % clear current event
if (eventnow==1) % arrival
arrival=timenow+exponentialrv(lam,1); % next arrival
b4arrival=time<arrival;
event=[event(b4arrival) 1 event(˜b4arrival)];
time=[time(b4arrival) arrival time(˜b4arrival)];
if n<c %call admitted
admits=admits+1;
n=n+1;
depart=timenow+exponentialrv(mu,1);
b4depart=time<depart;
event=[event(b4depart) -1 event(˜b4depart)];
time=[time(b4depart) depart time(˜b4depart)];
else
blocks=blocks+1; %one more block, immed departure
disp(sprintf(’Time %10.3d Admits %10d Blocks %10d’,...
timenow,admits,blocks));
end
elseif (eventnow==-1) %departure
n=n-1;
end
end
Figure 5: Discrete event simulation of the blocking switch of Quiz 10.13.
66](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-66-320.jpg)
![Quiz Solutions – Chapter 11
Quiz 11.1
By Theorem 11.2,
µY = µX
∞
−∞
h(t)dt = 2
∞
0
e−t
dt = 2 (1)
Since RX (τ) = δ(τ), the autocorrelation function of the output is
RY (τ) =
∞
−∞
h(u)
∞
−∞
h(v)δ(τ + u − v) dv du =
∞
−∞
h(u)h(τ + u) du (2)
For τ > 0, we have
RY (τ) =
∞
0
e−u
e−τ−u
du = e−τ
∞
0
e−2u
du =
1
2
e−τ
(3)
For τ < 0, we can deduce that RY (τ) = 1
2e−|τ| by symmetry. Just to be safe though, we
can double check. For τ < 0,
RY (τ) =
∞
−τ
h(u)h(τ + u) du =
∞
−τ
e−u
e−τ−u
du =
1
2
eτ
(4)
Hence,
RY (τ) =
1
2
e−|τ|
(5)
Quiz 11.2
The expected value of the output is
µY = µX
∞
n=−∞
hn = 0.5(1 + −1) = 0 (1)
The autocorrelation of the output is
RY [n] =
1
i=0
1
j=0
hi h j RX [n + i − j] (2)
= 2RX [n] − RX [n − 1] − RX [n + 1] =
1 n = 0
0 otherwise
(3)
Since µY = 0, The variance of Yn is Var[Yn] = E[Y2
n ] = RY [0] = 1.
67](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-67-320.jpg)
![−15 −10 −5 0 5 10 15
0
0.2
0.4
0.6
f
S
X
(f)
−1500−1000 −500 0 500 1000 1500
0
2
4
6
8
x 10
f
S
X
(f)
−0.2 −0.1 0 0.1 0.2
−5
0
5
10
τ
R
X
(τ)
−2 −1 0 1 2
x 10
−3
−5
0
5
10
τ
R
X
(τ)
(a) W = 10 (b) W = 1000
Figure 6: The autocorrelation RX (τ) and power spectral density SX ( f ) for process X(t) in
Quiz 11.5.
Quiz 11.3
By Theorem 11.8, Y = Y33 Y34 Y35 is a Gaussian random vector since Xn is
a Gaussian random process. Moreover, by Theorem 11.5, each Yn has expected value
E[Yn] = µX
∞
n=−∞ hn = 0. Thus E[Y] = 0. Fo find the PDF of the Gaussian vector
Y, we need to find the covariance matrix CY, which equals the correlation matrix RY since
Y has zero expected value. One way to find the RY is to observe that RY has the Toeplitz
structure of Theorem 11.6 and to use Theorem 11.5 to find the autocorrelation function
RY [n] =
∞
i=−∞
∞
j=−∞
hi h j RX [n + i − j]. (1)
Despite the fact that RX [k] is an impulse, using Equation (1) is surprisingly tedious because
we still need to sum over all i and j such that n + i − j = 0.
In this problem, it is simpler to observe that Y = HX where
X = X30 X31 X32 X33 X34 X35 (2)
and
H =
1
4
⎡
⎣
1 1 1 1 0 0
0 1 1 1 1 0
0 0 1 1 1 1
⎤
⎦ . (3)
In this case, following Theorem 11.7, or by directly applying Theorem 5.13 with µX = 0
and A = H, we obtain RY = HRXH . Since RX [n] = δn, RX = I, the identity matrix.
68](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-68-320.jpg)
![Thus
CY = RY = HH =
1
16
⎡
⎣
4 3 2
3 4 3
2 3 4
⎤
⎦ . (4)
It follows (very quickly if you use MATLAB for 3 × 3 matrix inversion) that
C−1
Y = 16
⎡
⎣
7/12 −1/2 1/12
−1/2 1 −1/2
1/12 −1/2 7/12
⎤
⎦ . (5)
Thus, the PDF of Y is
fY (y) =
1
(2π)3/2[det (CY)]1/2
exp −
1
2
y C−1
Y y . (6)
A disagreeable amount of algebra will show det(CY) = 3/1024 and that the PDF can be
“simplified” to
fY (y) =
16
√
6π3
exp −8
7
12
y2
33 + y2
34 +
7
12
y2
35 − y33y34 +
1
6
y33y35 − y34y35 . (7)
Equation (7) shows that one of the nicest features of the multivariate Gaussian distribution
is that y C−1
Y y is a very concise representation of the cross-terms in the exponent of fY(y).
Quiz 11.4
This quiz is solved using Theorem 11.9 for the case of k = 1 and M = 2. In this case,
Xn = Xn−1 Xn and
RXn =
RX [0] RX [1]
RX [1] RX [0]
=
1.1 0.9
0.9 1.1
(1)
and
RXn Xn+1 = E
Xn−1
Xn
Xn+1 =
RX [2]
RX [1]
=
0.81
0.9
. (2)
The MMSE linear first order filter for predicting Xn+1 at time n is the filter h such that
←−
h = R−1
Xn
RXn Xn+1 =
1.1 0.9
0.9 1.1
−1
0.81
0.9
=
1
400
81
261
. (3)
It follows that the filter is h = 261/400 81/400 and the MMSE linear predictor is
ˆXn+1 =
81
400
Xn−1 +
261
400
Xn. (4)
to find the mean square error, one approach is to follow the method of Example 11.13 and
to directly calculate
e∗
L = E (Xn+1 − ˆXn+1)2
. (5)
69](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-69-320.jpg)
![This method is workable for this simple problem but becomes increasingly tedious for
higher order filters. Instead, we can derive the mean square error for an arbitary prediction
filter h. Since ˆXn+1 =
←−
h Xn,
e∗
L = E Xn+1 −
←−
h Xn
2
(6)
= E (Xn+1 −
←−
h Xn)(Xn+1 −
←−
h Xn) (7)
= E (Xn+1 −
←−
h Xn)(Xn+1 − Xn
←−
h ) (8)
After a bit of algebra, we obtain
e∗
L = RX [0] − 2
←−
h RXn Xn+1 +
←−
h RXn
←−
h (9)
(10)
with the substitution
←−
h = R−1
Xn
RXn Xn+1, we obtain
e∗
L = RX [0] − RXn Xn+1
R−1
Xn
RXn Xn+1 (11)
= RX [0] −
←−
h RXn Xn+1 (12)
Note that this is essentially the same result as Theorem 9.7 with Y = Xn, X = Xn+1 and
ˆa =
←−
h . It is noteworthy that the result is derived in a much simpler way in the proof of
Theorem 9.7 by using the orthoginality property of the LMSE estimator.
In any case, the mean square error is
e∗
L = RX [0] −
←−
h RXn Xn+1 = 1.1 −
1
400
81 261
0.81
0.9
=
506
1451
= 0.3487. (13)
recalling that the blind estimate would yield a mean square error of Var[X] = 1.1, we see
that observing Xn−1 and Xn improves the accuracy of our prediction of Xn+1.
Quiz 11.5
(1) By Theorem 11.13(b), the average power of X(t) is
E X2
(t) =
∞
−∞
SX ( f ) d f =
W
−W
5
W
d f = 10 Watts (1)
(2) The autocorrelation function is the inverse Fourier transform of SX ( f ). Consulting
Table 11.1, we note that
SX ( f ) = 10
1
2W
rect
f
2W
(2)
It follows that the inverse transform of SX ( f ) is
RX (τ) = 10 sinc(2Wτ) = 10
sin(2πWτ)
2πWτ
(3)
(3) For W = 10 Hz and W = 1 kHZ, graphs of SX ( f ) and RX (τ) appear in Figure 6.
70](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-70-320.jpg)
![Quiz 11.6
In a sampled system, the discrete time impulse δ[n] has a flat discrete Fourier transform.
That is, if RX [n] = 10δ[n], then
SX (φ) =
∞
n=−∞
10δ[n]e− j2πφn
= 10 (1)
Thus, RX [n] = 10δ[n]. (This quiz is really lame!)
Quiz 11.7
Since Y(t) = X(t − t0),
RXY (t, τ) = E [X(t)Y(t + τ)] = E [X(t)X(t + τ − t0)] = RX (τ − t0) (1)
We see that RXY (t, τ) = RXY (τ) = RX (τ − t0). From Table 11.1, we recall the prop-
erty that g(τ − τ0) has Fourier transform G( f )e− j2π f τ0. Thus the Fourier transform of
RXY (τ) = RX (τ − t0) = g(τ − t0) is
SXY ( f ) = SX ( f )e− j2π f t0. (2)
Quiz 11.8
We solve this quiz using Theorem 11.17. First we need some preliminary facts. Let
a0 = 5,000 so that
RX (τ) =
1
a0
a0e−a0|τ|
. (1)
Consulting with the Fourier transforms in Table 11.1, we see that
SX ( f ) =
1
a0
2a2
0
a2
0 + (2π f )2
=
2a0
a2
0 + (2π f )2
(2)
The RC filter has impulse response h(t) = a1e−a1tu(t), where u(t) is the unit step function
and a1 = 1/RC where RC = 10−4 is the filter time constant. From Table 11.1,
H( f ) =
a1
a1 + j2π f
(3)
(1) Theorem 11.17,
SXY ( f ) = H( f )SX ( f ) =
2a0a1
[a1 + j2π f ] a2
0 + (2π f )2
. (4)
(2) Again by Theorem 11.17,
SY ( f ) = H∗
( f )SXY ( f ) = |H( f )|2
SX ( f ). (5)
71](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-71-320.jpg)
![Quiz 11.9
This quiz implements an example of Equations (11.146) and (11.147) for a system in
which we filter Y(t) = X(t) + N(t) to produce an optimal linear estimate of X(t). The
solution to this quiz is just to find the filter ˆH( f ) using Equation (11.146) and to calculate
the mean square error eL∗ using Equation (11.147).
Comment: Since the text omitted the derivations of Equations (11.146) and (11.147), we
note that Example 10.24 showed that
RY (τ) = RX (τ) + RN (τ), RY X (τ) = RX (τ). (1)
Taking Fourier transforms, it follows that
SY ( f ) = SX ( f ) + SN ( f ), SY X ( f ) = SX ( f ). (2)
Now we can go on to the quiz, at peace with the derivations.
(1) Since µN = 0, RN (0) = Var[N] = 1. This implies
RN (0) =
∞
−∞
SN ( f ) d f =
B
−B
N0 d f = 2N0 B (3)
Thus N0 = 1/(2B). Because the noise process N(t) has constant power RN (0) = 1,
decreasing the single-sided bandwidth B increases the power spectral density of the
noise over frequencies | f | < B.
(2) Since RX (τ) = sinc(2Wτ), where W = 5,000 Hz, we see from Table 11.1 that
SX ( f ) =
1
104
rect
f
104
. (4)
The noise power spectral density can be written as
SN ( f ) = N0 rect
f
2B
=
1
2B
rect
f
2B
, (5)
From Equation (11.146), the optimal filter is
ˆH( f ) =
SX ( f )
SX ( f ) + SN ( f )
=
1
104 rect f
104
1
104 rect f
104 + 1
2B rect f
2B
. (6)
73](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-73-320.jpg)
![(3) We produce the output ˆX(t) by passing the noisy signal Y(t) through the filter ˆH( f ).
From Equation (11.147), the mean square error of the estimate is
e∗
L =
∞
−∞
SX ( f )SN ( f )
SX ( f ) + SN ( f )
d f (7)
=
∞
−∞
1
104 rect f
104
1
2B rect f
2B
1
104 rect f
104 + 1
2B rect f
2B
d f. (8)
To evaluate the MSE e∗
L, we need to whether B ≤ W. Since the problem asks us to
find the largest possible B, let’s suppose B ≤ W. We can go back and consider the
case B > W later. When B ≤ W, the MSE is
e∗
L =
B
−B
1
104
1
2B
1
104 + 1
2B
d f =
1
104
1
104 + 1
2B
=
1
1 + 5,000
B
(9)
To obtain MSE e∗
L ≤ 0.05 requires B ≤ 5,000/19 = 263.16 Hz.
Although this completes the solution to the quiz, what is happening may not be obvious.
The noise power is always Var[N] = 1 Watt, for all values of B. As B is decreased, the PSD
SN ( f ) becomes increasingly tall, but only over a bandwidth B that is decreasing. Thus as
B descreases, the filter ˆH( f ) makes an increasingly deep and narrow notch at frequencies
| f | ≤ B. Two examples of the filter ˆH( f ) are shown in Figure 7. As B shrinks, the filter
suppresses less of the signal of X(t). The result is that the MSE goes down.
Finally, we note that we can choose B very large and also achieve MSE e∗
L = 0.05. In
particular, when B > W = 5000, SN ( f ) = 1/2B over frequencies | f | < W. In this case,
the Wiener filter ˆH( f ) is an ideal (flat) lowpass filter
ˆH( f ) =
⎧
⎨
⎩
1
104
1
104 + 1
2B
| f | < 5,000,
0 otherwise.
(10)
Thus increasing B spreads the constant 1 watt of power of N(t) over more bandwidth. The
Wiener filter removes the noise that is outside the band of the desired signal. The mean
square error is
e∗
L =
5000
−5000
1
104
1
2B
1
104 + 1
2B
d f =
1
2B
1
104 + 1
2B
=
1
B
5000 + 1
(11)
In this case, B ≥ 9.5 × 104 guarantees e∗
L ≤ 0.05.
Quiz 11.10
It is fairly straightforward to find SX (φ) and SY (φ). The only thing to keep in mind is
to use fftc to transform the autocorrelation RX [ f ] into the power spectral density SX (φ).
The following MATLAB program generates and plots the functions shown in Figure 8
74](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-74-320.jpg)
![−5000 −2000 0 2000 5000
0
0.5
1
f
H(f)
−5000 −2000 0 2000 5000
0
0.5
1
f
H(f)
B = 500 B = 2500
Figure 7: Wiener filter for Quiz 11.9.
%mquiz11.m
N=32;
rx=[2 4 2]; SX=fftc(rx,N); %autocorrelation and PSD
stem(0:N-1,abs(sx));
xlabel(’n’);ylabel(’S_X(n/N)’);
h2=0.5*[1 1]; H2=fft(h2,N); %impulse/filter response: M=2
SY2=SX.* ((abs(H2)).ˆ2);
figure; stem(0:N-1,abs(SY2)); %PSD of Y for M=2
xlabel(’n’);ylabel(’S_{Y_2}(n/N)’);
h10=0.1*ones(1,10); H10=fft(h10,N); %impulse/filter response: M=10
SY10=sx.*((abs(H10)).ˆ2);
figure; stem(0:N-1,abs(SY10));
xlabel(’n’);ylabel(’S_{Y_{10}}(n/N)’);
Relative to M = 2, when M = 10, the filter H(φ) filters out almost all of the high
frequency components of X(t). In the context of Example 11.26, the low pass moving
average filter for M = 10 removes the high frquency components and results in a filter
output that varies very slowly.
As an aside, note that the vectors SX, SY2 and SY10 in mquiz11 should all be real-
valued vectors. However, the finite numerical precision of MATLAB results in tiny imagi-
nary parts. Although these imaginary parts have no computational significance, they tend
to confuse the stem function. Hence, we generate stem plots of the magnitude of each
power spectral density.
75](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-75-320.jpg)
![2
0 1
0.9
0.1
11
P =
⎡
⎣
0.9 0.1 0
0 0 1
1 0 0
⎤
⎦ (1)
With π = π0 π1 π2 , the system of equations π = π P yields π1 = 0.1π0 and
π2 = π1. This implies
π0 + π1 + π2 = π0(1 + 0.1 + 0.1) = 1 (2)
It follows that the limiting state probabilities are
π0 = 5/6, π1 = 1/12, π2 = 1/12. (3)
Quiz 12.4
The communicating classes are
C1 = {0, 1} C2 = {2, 3} C3 = {4, 5, 6} (1)
The states in C1 and C3 are aperiodic. The states in C2 have period 2. Once the system
enters a state in C1, the class C1 is never left. Thus the states in C1 are recurrent. That
is, C1 is a recurrent class. Similarly, the states in C3 are recurrent. On the other hand, the
states in C2 are transient. Once the system exits C2, the states in C2 are never reentered.
Quiz 12.5
At any time t, the state n can take on the values 0, 1, 2, . . .. The state transition proba-
bilities are
Pn−1,n = P [K > n|K > n − 1] =
P [K > n]
P [K > n − 1]
(1)
Pn−1,0 = P [K = n|K > n − 1] =
P [K = n]
P [K > n − 1]
(2)
(3)
The Markov chain resembles
0 1
P K=2[ ]
P K=[ 1]
3 4
P K=4[ ]
2
P K=3[ ]
P K=5[ ]
1 1 1 1 1
…...
78](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-78-320.jpg)
![The stationary probabilities satisfy
π0 = π0 P [K = 1] + π1, (4)
π1 = π0 P [K = 2] + π2, (5)
...
πk−1 = π0 P [K = k] + πk, k = 1, 2, . . . (6)
From Equation (4), we obtain
π1 = π0 (1 − P [K = 1]) = π0 P [K > 1] (7)
Similarly, Equation (5) implies
π2 = π1 − π0 P [K = 2] = π0 (P [K > 1] − P [K = 2]) = π0 P [K > 2] (8)
This suggests that πk = π0 P[K > k]. We verify this pattern by showing that πk =
π0 P[K > k] satisfies Equation (6):
π0 P [K > k − 1] = π0 P [K = k] + π0 P [K > k] . (9)
When we apply ∞
k=0 πk = 1, we obtain π0
∞
n=0 P[K > k] = 1. From Problem 2.5.11,
we recall that ∞
k=0 P[K > k] = E[K]. This implies
πn =
P [K > n]
E [K]
(10)
This Markov chain models repeated random countdowns. The system state is the time until
the counter expires. When the counter expires, the system is in state 0, and we randomly
reset the counter to a new value K = k and then we count down k units of time. Since we
spend one unit of time in each state, including state 0, we have k − 1 units of time left after
the state 0 counter reset. If we have a random variable W such that the PMF of W satisfies
PW (n) = πn, then W has a discrete PMF representing the remaining time of the counter at
a time in the distant future.
Quiz 12.6
(1) By inspection, the number of transitions need to return to state 0 is always a multiple
of 2. Thus the period of state 0 is d = 2.
(2) To find the stationary probabilities, we solve the system of equations π = πP and
3
i=0 πi = 1:
π0 = (3/4)π1 + (1/4)π3 (1)
π1 = (1/4)π0 + (1/4)π2 (2)
π2 = (1/4)π1 + (3/4)π3 (3)
1 = π0 + π1 + π2 + π3 (4)
79](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-79-320.jpg)
![Solving the second and third equations for π2 and π3 yields
π2 = 4π1 − π0 π3 = (4/3)π2 − (1/3)π1 = 5π1 − (4/3)π0 (5)
Substituting π3 back into the first equation yields
π0 = (3/4)π1 + (1/4)π3 = (3/4)π1 + (5/4)π1 − (1/3)π0 (6)
This implies π1 = (2/3)π0. It follows from the first and second equations that
π2 = (5/3)π0 and π3 = 2π0. Lastly, we choose π0 so the state probabilities sum to
1:
1 = π0 + π1 + π2 + π3 = π0 1 +
2
3
+
5
3
+ 2 =
16
3
π0 (7)
It follows that the state probabilities are
π0 =
3
16
π1 =
2
16
π2 =
5
16
π3 =
6
16
(8)
(3) Since the system starts in state 0 at time 0, we can use Theorem 12.14 to find the
limiting probability that the system is in state 0 at time nd:
lim
n→∞
P00(nd) = dπ0 =
3
8
(9)
Quiz 12.7
The Markov chain has the same structure as that in Example 12.22. The only difference
is the modified transition rates:
0 1
1
3 4
( )2/3 a
1 - ( )2/3
a
( )3/4
a
1 - 3/4( )
a
( )4/5
a
1 - 4/5( )a
2
( )1/2 a
1- 1/2( )
a
…
The event T00 > n occurs if the system reaches state n before returning to state 0, which
occurs with probability
P [T00 > n] = 1 ×
1
2
α
×
2
3
α
× · · · ×
n − 1
n
α
=
1
n
α
. (1)
Thus the CDF of T00 satisfies FT00(n) = 1−P[T00 > n] = 1−1/nα. To determine whether
state 0 is recurrent, we observe that for all α > 0
P [V00] = lim
n→∞
FT00 (n) = lim
n→∞
1 −
1
nα
= 1. (2)
80](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-80-320.jpg)
![Thus state 0 is recurrent for all α > 0. Since the chain has only one communicating class,
all states are recurrent. ( We also note that if α = 0, then all states are transient.)
To determine whether the chain is null recurrent or positive recurrent, we need to calcu-
late E[T00]. In Example 12.24, we did this by deriving the PMF PT00(n). In this problem,
it will be simpler to use the result of Problem 2.5.11 which says that ∞
k=0 P[K > k] =
E[K] for any non-negative integer-valued random variable K. Applying this result, the
expected time to return to state 0 is
E [T00] =
∞
n=0
P [T00 > n] = 1 +
∞
n=1
1
nα
. (3)
For 0 < α ≤ 1, 1/nα ≥ 1/n and it follows that
E [T00] ≥ 1 +
∞
n=1
1
n
= ∞. (4)
We conclude that the Markov chain is null recurrent for 0 < α ≤ 1. On the other hand, for
α > 1,
E [T00] = 2 +
∞
n=2
1
nα
. (5)
Note that for all n ≥ 2
1
nα
≤
n
n−1
dx
xα
(6)
This implies
E [T00] ≤ 2 +
∞
n=2
n
n−1
dx
xα
(7)
= 2 +
∞
1
dx
xα
(8)
= 2 +
x−α+1
−α + 1
∞
1
= 2 +
1
α − 1
< ∞ (9)
Thus for all α > 1, the Markov chain is positive recurrent.
Quiz 12.8
The number of customers in the ”friendly” store is given by the Markov chain
1 i i+1
p p p
( )( )1-p 1-q ( )( )1-p 1-q ( )( )1-p 1-q ( )( )1-p 1-q
( )1-p q ( )1-p q ( )1-p q ( )1-p q
0 ××× ×××
81](https://image.slidesharecdn.com/solucionesquiz-161205130657/85/Soluciones-quiz-81-320.jpg)
Soluciones quiz
- 1. Probability and Stochastic Processes A Friendly Introduction for Electrical and Computer Engineers Second Edition Quiz Solutions Roy D. Yates and David J. Goodman May 22, 2004 • The MATLAB section quizzes at the end of each chapter use programs available for download as the archive matcode.zip. This archive has programs of general pur- pose programs for solving probability problems as well as specific .m files associated with examples or quizzes in the text. Also available is a manual probmatlab.pdf describing the general purpose .m files in matcode.zip. • We have made a substantial effort to check the solution to every quiz. Nevertheless, there is a nonzero probability (in fact, a probability close to unity) that errors will be found. If you find errors or have suggestions or comments, please send email to ryates@winlab.rutgers.edu. When errors are found, corrected solutions will be posted at the website. 1
- 2. Quiz Solutions – Chapter 1 Quiz 1.1 In the Venn diagrams for parts (a)-(g) below, the shaded area represents the indicated set. M O T M O T M O T (1) R = T c (2) M ∪ O (3) M ∩ O M O T M O T M O T (4) R ∪ M (4) R ∩ M (6) T c − M Quiz 1.2 (1) A1 = {vvv, vvd, vdv, vdd} (2) B1 = {dvv, dvd, ddv, ddd} (3) A2 = {vvv, vvd, dvv, dvd} (4) B2 = {vdv, vdd, ddv, ddd} (5) A3 = {vvv, ddd} (6) B3 = {vdv, dvd} (7) A4 = {vvv, vvd, vdv, dvv, vdd, dvd, ddv} (8) B4 = {ddd, ddv, dvd, vdd} Recall that Ai and Bi are collectively exhaustive if Ai ∪ Bi = S. Also, Ai and Bi are mutually exclusive if Ai ∩ Bi = φ. Since we have written down each pair Ai and Bi above, we can simply check for these properties. The pair A1 and B1 are mutually exclusive and collectively exhaustive. The pair A2 and B2 are mutually exclusive and collectively exhaustive. The pair A3 and B3 are mutually exclusive but not collectively exhaustive. The pair A4 and B4 are not mutually exclusive since dvd belongs to A4 and B4. However, A4 and B4 are collectively exhaustive. 2
- 3. Quiz 1.3 There are exactly 50 equally likely outcomes: s51 through s100. Each of these outcomes has probability 0.02. (1) P[{s79}] = 0.02 (2) P[{s100}] = 0.02 (3) P[A] = P[{s90, . . . , s100}] = 11 × 0.02 = 0.22 (4) P[F] = P[{s51, . . . , s59}] = 9 × 0.02 = 0.18 (5) P[T ≥ 80] = P[{s80, . . . , s100}] = 21 × 0.02 = 0.42 (6) P[T < 90] = P[{s51, s52, . . . , s89}] = 39 × 0.02 = 0.78 (7) P[a C grade or better] = P[{s70, . . . , s100}] = 31 × 0.02 = 0.62 (8) P[student passes] = P[{s60, . . . , s100}] = 41 × 0.02 = 0.82 Quiz 1.4 We can describe this experiment by the event space consisting of the four possible events V B, V L, DB, and DL. We represent these events in the table: V D L 0.35 ? B ? ? In a roundabout way, the problem statement tells us how to fill in the table. In particular, P [V ] = 0.7 = P [V L] + P [V B] (1) P [L] = 0.6 = P [V L] + P [DL] (2) Since P[V L] = 0.35, we can conclude that P[V B] = 0.35 and that P[DL] = 0.6 − 0.35 = 0.25. This allows us to fill in two more table entries: V D L 0.35 0.25 B 0.35 ? The remaining table entry is filled in by observing that the probabilities must sum to 1. This implies P[DB] = 0.05 and the complete table is V D L 0.35 0.25 B 0.35 0.05 Finding the various probabilities is now straightforward: 3
- 4. (1) P[DL] = 0.25 (2) P[D ∪ L] = P[V L] + P[DL] + P[DB] = 0.35 + 0.25 + 0.05 = 0.65. (3) P[V B] = 0.35 (4) P[V ∪ L] = P[V ] + P[L] − P[V L] = 0.7 + 0.6 − 0.35 = 0.95 (5) P[V ∪ D] = P[S] = 1 (6) P[L B] = P[LLc] = 0 Quiz 1.5 (1) The probability of exactly two voice calls is P [NV = 2] = P [{vvd, vdv, dvv}] = 0.3 (1) (2) The probability of at least one voice call is P [NV ≥ 1] = P [{vdd, dvd, ddv, vvd, vdv, dvv, vvv}] (2) = 6(0.1) + 0.2 = 0.8 (3) An easier way to get the same answer is to observe that P [NV ≥ 1] = 1 − P [NV < 1] = 1 − P [NV = 0] = 1 − P [{ddd}] = 0.8 (4) (3) The conditional probability of two voice calls followed by a data call given that there were two voice calls is P [{vvd} |NV = 2] = P [{vvd} , NV = 2] P [NV = 2] = P [{vvd}] P [NV = 2] = 0.1 0.3 = 1 3 (5) (4) The conditional probability of two data calls followed by a voice call given there were two voice calls is P [{ddv} |NV = 2] = P [{ddv} , NV = 2] P [NV = 2] = 0 (6) The joint event of the outcome ddv and exactly two voice calls has probability zero since there is only one voice call in the outcome ddv. (5) The conditional probability of exactly two voice calls given at least one voice call is P [NV = 2|Nv ≥ 1] = P [NV = 2, NV ≥ 1] P [NV ≥ 1] = P [NV = 2] P [NV ≥ 1] = 0.3 0.8 = 3 8 (7) (6) The conditional probability of at least one voice call given there were exactly two voice calls is P [NV ≥ 1|NV = 2] = P [NV ≥ 1, NV = 2] P [NV = 2] = P [NV = 2] P [NV = 2] = 1 (8) Given that there were two voice calls, there must have been at least one voice call. 4
- 5. Quiz 1.6 In this experiment, there are four outcomes with probabilities P[{vv}] = (0.8)2 = 0.64 P[{vd}] = (0.8)(0.2) = 0.16 P[{dv}] = (0.2)(0.8) = 0.16 P[{dd}] = (0.2)2 = 0.04 When checking the independence of any two events A and B, it’s wise to avoid intuition and simply check whether P[AB] = P[A]P[B]. Using the probabilities of the outcomes, we now can test for the independence of events. (1) First, we calculate the probability of the joint event: P [NV = 2, NV ≥ 1] = P [NV = 2] = P [{vv}] = 0.64 (1) Next, we observe that P [NV ≥ 1] = P [{vd, dv, vv}] = 0.96 (2) Finally, we make the comparison P [NV = 2] P [NV ≥ 1] = (0.64)(0.96) = P [NV = 2, NV ≥ 1] (3) which shows the two events are dependent. (2) The probability of the joint event is P [NV ≥ 1, C1 = v] = P [{vd, vv}] = 0.80 (4) From part (a), P[NV ≥ 1] = 0.96. Further, P[C1 = v] = 0.8 so that P [NV ≥ 1] P [C1 = v] = (0.96)(0.8) = 0.768 = P [NV ≥ 1, C1 = v] (5) Hence, the events are dependent. (3) The problem statement that the calls were independent implies that the events the second call is a voice call, {C2 = v}, and the first call is a data call, {C1 = d} are independent events. Just to be sure, we can do the calculations to check: P [C1 = d, C2 = v] = P [{dv}] = 0.16 (6) Since P[C1 = d]P[C2 = v] = (0.2)(0.8) = 0.16, we confirm that the events are independent. Note that this shouldn’t be surprising since we used the information that the calls were independent in the problem statement to determine the probabilities of the outcomes. (4) The probability of the joint event is P [C2 = v, NV is even] = P [{vv}] = 0.64 (7) Also, each event has probability P [C2 = v] = P [{dv, vv}] = 0.8, P [NV is even] = P [{dd, vv}] = 0.68 (8) Thus, P[C2 = v]P[NV is even] = (0.8)(0.68) = 0.544. Since P[C2 = v, NV is even] = 0.544, the events are dependent. 5
- 6. Quiz 1.7 Let Fi denote the event that that the user is found on page i. The tree for the experiment is ¨¨¨¨¨¨ F10.8 Fc 10.2 ¨¨¨¨¨¨ F20.8 Fc 20.2 ¨¨¨¨¨¨ F30.8 Fc 30.2 The user is found unless all three paging attempts fail. Thus the probability the user is found is P [F] = 1 − P Fc 1 Fc 2 Fc 3 = 1 − (0.2)3 = 0.992 (1) Quiz 1.8 (1) We can view choosing each bit in the code word as a subexperiment. Each subex- periment has two possible outcomes: 0 and 1. Thus by the fundamental principle of counting, there are 2 × 2 × 2 × 2 = 24 = 16 possible code words. (2) An experiment that can yield all possible code words with two zeroes is to choose which 2 bits (out of 4 bits) will be zero. The other two bits then must be ones. There are 4 2 = 6 ways to do this. Hence, there are six code words with exactly two zeroes. For this problem, it is also possible to simply enumerate the six code words: 1100, 1010, 1001, 0101, 0110, 0011. (3) When the first bit must be a zero, then the first subexperiment of choosing the first bit has only one outcome. For each of the next three bits, we have two choices. In this case, there are 1 × 2 × 2 × 2 = 8 ways of choosing a code word. (4) For the constant ratio code, we can specify a code word by choosing M of the bits to be ones. The other N − M bits will be zeroes. The number of ways of choosing such a code word is N M . For N = 8 and M = 3, there are 8 3 = 56 code words. Quiz 1.9 (1) In this problem, k bits received in error is the same as k failures in 100 trials. The failure probability is = 1 − p and the success probability is 1 − = p. That is, the probability of k bits in error and 100 − k correctly received bits is P Sk,100−k = 100 k k (1 − )100−k (1) 6
- 7. For = 0.01, P S0,100 = (1 − )100 = (0.99)100 = 0.3660 (2) P S1,99 = 100(0.01)(0.99)99 = 0.3700 (3) P S2,98 = 4950(0.01)2 (0.99)9 8 = 0.1849 (4) P S3,97 = 161, 700(0.01)3 (0.99)97 = 0.0610 (5) (2) The probability a packet is decoded correctly is just P [C] = P S0,100 + P S1,99 + P S2,98 + P S3,97 = 0.9819 (6) Quiz 1.10 Since the chip works only if all n transistors work, the transistors in the chip are like devices in series. The probability that a chip works is P[C] = pn. The module works if either 8 chips work or 9 chips work. Let Ck denote the event that exactly k chips work. Since transistor failures are independent of each other, chip failures are also independent. Thus each P[Ck] has the binomial probability P [C8] = 9 8 (P [C])8 (1 − P [C])9−8 = 9p8n (1 − pn ), (1) P [C9] = (P [C])9 = p9n . (2) The probability a memory module works is P [M] = P [C8] + P [C9] = p8n (9 − 8pn ) (3) Quiz 1.11 R=rand(1,100); X=(R<= 0.4) ... + (2*(R>0.4).*(R<=0.9)) ... + (3*(R>0.9)); Y=hist(X,1:3) For a MATLAB simulation, we first gen- erate a vector R of 100 random numbers. Second, we generate vector X as a func- tion of R to represent the 3 possible out- comes of a flip. That is, X(i)=1 if flip i was heads, X(i)=2 if flip i was tails, and X(i)=3) is flip i landed on the edge. To see how this works, we note there are three cases: • If R(i) <= 0.4, then X(i)=1. • If 0.4 < R(i) and R(i)<=0.9, then X(i)=2. • If 0.9 < R(i), then X(i)=3. These three cases will have probabilities 0.4, 0.5 and 0.1. Lastly, we use the hist function to count how many occurences of each possible value of X(i). 7
- 8. Quiz Solutions – Chapter 2 Quiz 2.1 The sample space, probabilities and corresponding grades for the experiment are Outcome P[·] G BB 0.36 3.0 BC 0.24 2.5 C B 0.24 2.5 CC 0.16 2 Quiz 2.2 (1) To find c, we recall that the PMF must sum to 1. That is, 3 n=1 PN (n) = c 1 + 1 2 + 1 3 = 1 (1) This implies c = 6/11. Now that we have found c, the remaining parts are straight- forward. (2) P[N = 1] = PN (1) = c = 6/11 (3) P[N ≥ 2] = PN (2) + PN (3) = c/2 + c/3 = 5/11 (4) P[N > 3] = ∞ n=4 PN (n) = 0 Quiz 2.3 Decoding each transmitted bit is an independent trial where we call a bit error a “suc- cess.” Each bit is in error, that is, the trial is a success, with probability p. Now we can interpret each experiment in the generic context of independent trials. (1) The random variable X is the number of trials up to and including the first success. Similar to Example 2.11, X has the geometric PMF PX (x) = p(1 − p)x−1 x = 1, 2, . . . 0 otherwise (1) (2) If p = 0.1, then the probability exactly 10 bits are sent is P [X = 10] = PX (10) = (0.1)(0.9)9 = 0.0387 (2) 8
- 9. The probability that at least 10 bits are sent is P[X ≥ 10] = ∞ x=10 PX (x). This sum is not too hard to calculate. However, its even easier to observe that X ≥ 10 if the first 10 bits are transmitted correctly. That is, P [X ≥ 10] = P [first 10 bits are correct] = (1 − p)10 (3) For p = 0.1, P[X ≥ 10] = 0.910 = 0.3487. (3) The random variable Y is the number of successes in 100 independent trials. Just as in Example 2.13, Y has the binomial PMF PY (y) = 100 y py (1 − p)100−y (4) If p = 0.01, the probability of exactly 2 errors is P [Y = 2] = PY (2) = 100 2 (0.01)2 (0.99)98 = 0.1849 (5) (4) The probability of no more than 2 errors is P [Y ≤ 2] = PY (0) + PY (1) + PY (2) (6) = (0.99)100 + 100(0.01)(0.99)99 + 100 2 (0.01)2 (0.99)98 (7) = 0.9207 (8) (5) Random variable Z is the number of trials up to and including the third success. Thus Z has the Pascal PMF (see Example 2.15) PZ (z) = z − 1 2 p3 (1 − p)z−3 (9) Note that PZ (z) > 0 for z = 3, 4, 5, . . .. (6) If p = 0.25, the probability that the third error occurs on bit 12 is PZ (12) = 11 2 (0.25)3 (0.75)9 = 0.0645 (10) Quiz 2.4 Each of these probabilities can be read off the CDF FY (y). However, we must keep in mind that when FY (y) has a discontinuity at y0, FY (y) takes the upper value FY (y+ 0 ). (1) P[Y < 1] = FY (1−) = 0 9
- 10. (2) P[Y ≤ 1] = FY (1) = 0.6 (3) P[Y > 2] = 1 − P[Y ≤ 2] = 1 − FY (2) = 1 − 0.8 = 0.2 (4) P[Y ≥ 2] = 1 − P[Y < 2] = 1 − FY (2−) = 1 − 0.6 = 0.4 (5) P[Y = 1] = P[Y ≤ 1] − P[Y < 1] = FY (1+) − FY (1−) = 0.6 (6) P[Y = 3] = P[Y ≤ 3] − P[Y < 3] = FY (3+) − FY (3−) = 0.8 − 0.8 = 0 Quiz 2.5 (1) With probability 0.7, a call is a voice call and C = 25. Otherwise, with probability 0.3, we have a data call and C = 40. This corresponds to the PMF PC (c) = ⎧ ⎨ ⎩ 0.7 c = 25 0.3 c = 40 0 otherwise (1) (2) The expected value of C is E [C] = 25(0.7) + 40(0.3) = 29.5 cents (2) Quiz 2.6 (1) As a function of N, the cost T is T = 25N + 40(3 − N) = 120 − 15N (1) (2) To find the PMF of T , we can draw the following tree: ¨¨¨ ¨¨¨¨N=0 0.1 rrrrr rrN=3 0.3 $$$$$$$N=10.3 ˆˆˆˆˆˆˆN=20.3 •T =120 •T =105 •T =90 •T =75 From the tree, we can write down the PMF of T : PT (t) = ⎧ ⎨ ⎩ 0.3 t = 75, 90, 105 0.1 t = 120 0 otherwise (2) From the PMF PT (t), the expected value of T is E [T ] = 75PT (75) + 90PT (90) + 105PT (105) + 120PT (120) (3) = (75 + 90 + 105)(0.3) + 120(0.1) = 62 (4) 10
- 11. Quiz 2.7 (1) Using Definition 2.14, the expected number of applications is E [A] = 4 a=1 aPA (a) = 1(0.4) + 2(0.3) + 3(0.2) + 4(0.1) = 2 (1) (2) The number of memory chips is M = g(A) where g(A) = ⎧ ⎨ ⎩ 4 A = 1, 2 6 A = 3 8 A = 4 (2) (3) By Theorem 2.10, the expected number of memory chips is E [M] = 4 a=1 g(A)PA (a) = 4(0.4) + 4(0.3) + 6(0.2) + 8(0.1) = 4.8 (3) Since E[A] = 2, g(E[A]) = g(2) = 4. However, E[M] = 4.8 = g(E[A]). The two quantities are different because g(A) is not of the form αA + β. Quiz 2.8 The PMF PN (n) allows to calculate each of the desired quantities. (1) The expected value of N is E [N] = 2 n=0 nPN (n) = 0(0.1) + 1(0.4) + 2(0.5) = 1.4 (1) (2) The second moment of N is E N2 = 2 n=0 n2 PN (n) = 02 (0.1) + 12 (0.4) + 22 (0.5) = 2.4 (2) (3) The variance of N is Var[N] = E N2 − (E [N])2 = 2.4 − (1.4)2 = 0.44 (3) (4) The standard deviation is σN = √ Var[N] = √ 0.44 = 0.663. 11
- 12. Quiz 2.9 (1) From the problem statement, we learn that the conditional PMF of N given the event I is PN|I (n) = 0.02 n = 1, 2, . . . , 50 0 otherwise (1) (2) Also from the problem statement, the conditional PMF of N given the event T is PN|T (n) = 0.2 n = 1, 2, 3, 4, 5 0 otherwise (2) (3) The problem statement tells us that P[T ] = 1 − P[I] = 3/4. From Theorem 1.10 (the law of total probability), we find the PMF of N is PN (n) = PN|T (n) P [T ] + PN|I (n) P [I] (3) = ⎧ ⎨ ⎩ 0.2(0.75) + 0.02(0.25) n = 1, 2, 3, 4, 5 0(0.75) + 0.02(0.25) n = 6, 7, . . . , 50 0 otherwise (4) = ⎧ ⎨ ⎩ 0.155 n = 1, 2, 3, 4, 5 0.005 n = 6, 7, . . . , 50 0 otherwise (5) (4) First we find P [N ≤ 10] = 10 n=1 PN (n) = (0.155)(5) + (0.005)(5) = 0.80 (6) By Theorem 2.17, the conditional PMF of N given N ≤ 10 is PN|N≤10 (n) = PN (n) P[N≤10] n ≤ 10 0 otherwise (7) = ⎧ ⎨ ⎩ 0.155/0.8 n = 1, 2, 3, 4, 5 0.005/0.8 n = 6, 7, 8, 9, 10 0 otherwise (8) = ⎧ ⎨ ⎩ 0.19375 n = 1, 2, 3, 4, 5 0.00625 n = 6, 7, 8, 9, 10 0 otherwise (9) (5) Once we have the conditional PMF, calculating conditional expectations is easy. E [N|N ≤ 10] = n nPN|N≤10 (n) (10) = 5 n=1 n(0.19375) + 10 n=6 n(0.00625) (11) = 3.15625 (12) 12
- 13. 0 50 100 0 2 4 6 8 10 0 500 1000 0 2 4 6 8 10 (a) samplemean(100) (b) samplemean(1000) Figure 1: Two examples of the output of samplemean(k) (6) To find the conditional variance, we first find the conditional second moment E N2 |N ≤ 10 = n n2 PN|N≤10 (n) (13) = 5 n=1 n2 (0.19375) + 10 n=6 n2 (0.00625) (14) = 55(0.19375) + 330(0.00625) = 12.71875 (15) The conditional variance is Var[N|N ≤ 10] = E N2 |N ≤ 10 − (E [N|N ≤ 10])2 (16) = 12.71875 − (3.15625)2 = 2.75684 (17) Quiz 2.10 The function samplemean(k) generates and plots five mn sequences for n = 1, 2, . . . , k. The ith column M(:,i) of M holds a sequence m1, m2, . . . , mk. function M=samplemean(k); K=(1:k)’; M=zeros(k,5); for i=1:5, X=duniformrv(0,10,k); M(:,i)=cumsum(X)./K; end; plot(K,M); Examples of the function calls (a) samplemean(100) and (b) samplemean(1000) are shown in Figure 1. Each time samplemean(k) is called produces a random output. What is observed in these figures is that for small n, mn is fairly random but as n gets 13
- 14. large, mn gets close to E[X] = 5. Although each sequence m1, m2, . . . that we generate is random, the sequences always converges to E[X]. This random convergence is analyzed in Chapter 7. 14
- 15. Quiz Solutions – Chapter 3 Quiz 3.1 The CDF of Y is 0 2 4 0 0.5 1 y F Y (y) FY (y) = ⎧ ⎨ ⎩ 0 y < 0 y/4 0 ≤ y ≤ 4 1 y > 4 (1) From the CDF FY (y), we can calculate the probabilities: (1) P[Y ≤ −1] = FY (−1) = 0 (2) P[Y ≤ 1] = FY (1) = 1/4 (3) P[2 < Y ≤ 3] = FY (3) − FY (2) = 3/4 − 2/4 = 1/4 (4) P[Y > 1.5] = 1 − P[Y ≤ 1.5] = 1 − FY (1.5) = 1 − (1.5)/4 = 5/8 Quiz 3.2 (1) First we will find the constant c and then we will sketch the PDF. To find c, we use the fact that ∞ −∞ fX (x) dx = 1. We will evaluate this integral using integration by parts: ∞ −∞ fX (x) dx = ∞ 0 cxe−x/2 dx (1) = −2cxe−x/2 ∞ 0 =0 + ∞ 0 2ce−x/2 dx (2) = −4ce−x/2 ∞ 0 = 4c (3) Thus c = 1/4 and X has the Erlang (n = 2, λ = 1/2) PDF 0 5 10 15 0 0.1 0.2 x f X (x) fX (x) = (x/4)e−x/2 x ≥ 0 0 otherwise (4) 15
- 16. (2) To find the CDF FX (x), we first note X is a nonnegative random variable so that FX (x) = 0 for all x < 0. For x ≥ 0, FX (x) = x 0 fX (y) dy = x 0 y 4 e−y/2 dy (5) = − y 2 e−y/2 x 0 − x 0 − 1 2 e−y/2 dy (6) = 1 − x 2 e−x/2 − e−x/2 (7) The complete expression for the CDF is 0 5 10 15 0 0.5 1 x F X (x) FX (x) = 1 − x 2 + 1 e−x/2 x ≥ 0 0 otherwise (8) (3) From the CDF FX (x), P [0 ≤ X ≤ 4] = FX (4) − FX (0) = 1 − 3e−2 . (9) (4) Similarly, P [−2 ≤ X ≤ 2] = FX (2) − FX (−2) = 1 − 3e−1 . (10) Quiz 3.3 The PDF of Y is −2 0 2 0 1 2 3 y f Y (y) fY (y) = 3y2/2 −1 ≤ y ≤ 1, 0 otherwise. (1) (1) The expected value of Y is E [Y] = ∞ −∞ y fY (y) dy = 1 −1 (3/2)y3 dy = (3/8)y4 1 −1 = 0. (2) Note that the above calculation wasn’t really necessary because E[Y] = 0 whenever the PDF fY (y) is an even function (i.e., fY (y) = fY (−y)). (2) The second moment of Y is E Y2 = ∞ −∞ y2 fY (y) dy = 1 −1 (3/2)y4 dy = (3/10)y5 1 −1 = 3/5. (3) 16
- 17. (3) The variance of Y is Var[Y] = E Y2 − (E [Y])2 = 3/5. (4) (4) The standard deviation of Y is σY = √ Var[Y] = √ 3/5. Quiz 3.4 (1) When X is an exponential (λ) random variable, E[X] = 1/λ and Var[X] = 1/λ2. Since E[X] = 3 and Var[X] = 9, we must have λ = 1/3. The PDF of X is fX (x) = (1/3)e−x/3 x ≥ 0, 0 otherwise. (1) (2) We know X is a uniform (a, b) random variable. To find a and b, we apply Theo- rem 3.6 to write E [X] = a + b 2 = 3 Var[X] = (b − a)2 12 = 9. (2) This implies a + b = 6, b − a = ±6 √ 3. (3) The only valid solution with a < b is a = 3 − 3 √ 3, b = 3 + 3 √ 3. (4) The complete expression for the PDF of X is fX (x) = 1/(6 √ 3) 3 − 3 √ 3 ≤ x < 3 + 3 √ 3, 0 otherwise. (5) Quiz 3.5 Each of the requested probabilities can be calculated using (z) function and Table 3.1 or Q(z) and Table 3.2. We start with the sketches. (1) The PDFs of X and Y are shown below. The fact that Y has twice the standard deviation of X is reflected in the greater spread of fY (y). However, it is important to remember that as the standard deviation increases, the peak value of the Gaussian PDF goes down. −5 0 5 0 0.2 0.4 x y f X (x)f Y (y) ← f X (x) ← f Y (y) 17
- 18. (2) Since X is Gaussian (0, 1), P [−1 < X ≤ 1] = FX (1) − FX (−1) (1) = (1) − (−1) = 2 (1) − 1 = 0.6826. (2) (3) Since Y is Gaussian (0, 2), P [−1 < Y ≤ 1] = FY (1) − FY (−1) (3) = 1 σY − −1 σY = 2 1 2 − 1 = 0.383. (4) (4) Again, since X is Gaussian (0, 1), P[X > 3.5] = Q(3.5) = 2.33 × 10−4. (5) Since Y is Gaussian (0, 2), P[Y > 3.5] = Q(3.5 2 ) = Q(1.75) = 1 − (1.75) = 0.0401. Quiz 3.6 The CDF of X is −2 0 2 0 0.5 1 x F X (x) FX (x) = ⎧ ⎨ ⎩ 0 x < −1, (x + 1)/4 −1 ≤ x < 1, 1 x ≥ 1. (1) The following probabilities can be read directly from the CDF: (1) P[X ≤ 1] = FX (1) = 1. (2) P[X < 1] = FX (1−) = 1/2. (3) P[X = 1] = FX (1+) − FX (1−) = 1 − 1/2 = 1/2. (4) We find the PDF fY (y) by taking the derivative of FY (y). The resulting PDF is −2 0 2 0 0.5 x f X (x) 0.5 fX (x) = ⎧ ⎨ ⎩ 1/4 −1 ≤ x < 1, (1/2)δ(x − 1) x = 1, 0 otherwise. (2) Quiz 3.7 18
- 19. (1) Since X is always nonnegative, FX (x) = 0 for x < 0. Also, FX (x) = 1 for x ≥ 2 since its always true that x ≤ 2. Lastly, for 0 ≤ x ≤ 2, FX (x) = x −∞ fX (y) dy = x 0 (1 − y/2) dy = x − x2 /4. (1) The complete CDF of X is −1 0 1 2 3 0 0.5 1 x F X (x) FX (x) = ⎧ ⎨ ⎩ 0 x < 0, x − x2/4 0 ≤ x ≤ 2, 1 x > 2. (2) (2) The probability that Y = 1 is P [Y = 1] = P [X ≥ 1] = 1 − FX (1) = 1 − 3/4 = 1/4. (3) (3) Since X is nonnegative, Y is also nonnegative. Thus FY (y) = 0 for y < 0. Also, because Y ≤ 1, FY (y) = 1 for all y ≥ 1. Finally, for 0 < y < 1, FY (y) = P [Y ≤ y] = P [X ≤ y] = FX (y) . (4) Using the CDF FX (x), the complete expression for the CDF of Y is −1 0 1 2 3 0 0.5 1 y F Y (y) FY (y) = ⎧ ⎨ ⎩ 0 y < 0, y − y2/4 0 ≤ y < 1, 1 y ≥ 1. (5) As expected, we see that the jump in FY (y) at y = 1 is exactly equal to P[Y = 1]. (4) By taking the derivative of FY (y), we obtain the PDF fY (y). Note that when y < 0 or y > 1, the PDF is zero. −1 0 1 2 3 0 0.5 1 1.5 y f Y (y) 0.25 fY (y) = 1 − y/2 + (1/4)δ(y − 1) 0 ≤ y ≤ 1 0 otherwise (6) Quiz 3.8 (1) P[Y ≤ 6] = 6 −∞ fY (y) dy = 6 0 (1/10) dy = 0.6 . 19
- 20. (2) From Definition 3.15, the conditional PDF of Y given Y ≤ 6 is fY|Y≤6 (y) = fY (y) P[Y≤6] y ≤ 6, 0 otherwise, = 1/6 0 ≤ y ≤ 6, 0 otherwise. (1) (3) The probability Y > 8 is P [Y > 8] = 10 8 1 10 dy = 0.2 . (2) (4) From Definition 3.15, the conditional PDF of Y given Y > 8 is fY|Y>8 (y) = fY (y) P[Y>8] y > 8, 0 otherwise, = 1/2 8 < y ≤ 10, 0 otherwise. (3) (5) From the conditional PDF fY|Y≤6(y), we can calculate the conditional expectation E [Y|Y ≤ 6] = ∞ −∞ y fY|Y≤6 (y) dy = 6 0 y 6 dy = 3. (4) (6) From the conditional PDF fY|Y>8(y), we can calculate the conditional expectation E [Y|Y > 8] = ∞ −∞ yfY|Y>8 (y) dy = 10 8 y 2 dy = 9. (5) Quiz 3.9 A natural way to produce random variables with PDF fT|T >2(t) is to generate samples of T with PDF fT (t) and then to discard those samples which fail to satisfy the condition T > 2. Here is a MATLAB function that uses this method: function t=t2rv(m) i=0;lambda=1/3; t=zeros(m,1); while (i<m), x=exponentialrv(lambda,1); if (x>2) t(i+1)=x; i=i+1; end end A second method exploits the fact that if T is an exponential (λ) random variable, then T = T + 2 has PDF fT (t) = fT|T >2(t). In this case the command t=2.0+exponentialrv(1/3,m) generates the vector t. 20
- 21. Quiz Solutions – Chapter 4 Quiz 4.1 Each value of the joint CDF can be found by considering the corresponding probability. (1) FX,Y (−∞, 2) = P[X ≤ −∞, Y ≤ 2] ≤ P[X ≤ −∞] = 0 since X cannot take on the value −∞. (2) FX,Y (∞, ∞) = P[X ≤ ∞, Y ≤ ∞] = 1. This result is given in Theorem 4.1. (3) FX,Y (∞, y) = P[X ≤ ∞, Y ≤ y] = P[Y ≤ y] = FY (y). (4) FX,Y (∞, −∞) = P[X ≤ ∞, Y ≤ −∞] = 0 since Y cannot take on the value −∞. Quiz 4.2 From the joint PMF of Q and G given in the table, we can calculate the requested probabilities by summing the PMF over those values of Q and G that correspond to the event. (1) The probability that Q = 0 is P [Q = 0] = PQ,G (0, 0) + PQ,G (0, 1) + PQ,G (0, 2) + PQ,G (0, 3) (1) = 0.06 + 0.18 + 0.24 + 0.12 = 0.6 (2) (2) The probability that Q = G is P [Q = G] = PQ,G (0, 0) + PQ,G (1, 1) = 0.18 (3) (3) The probability that G > 1 is P [G > 1] = 3 g=2 1 q=0 PQ,G (q, g) (4) = 0.24 + 0.16 + 0.12 + 0.08 = 0.6 (5) (4) The probability that G > Q is P [G > Q] = 1 q=0 3 g=q+1 PQ,G (q, g) (6) = 0.18 + 0.24 + 0.12 + 0.16 + 0.08 = 0.78 (7) 21
- 22. Quiz 4.3 By Theorem 4.3, the marginal PMF of H is PH (h) = b=0,2,4 PH,B (h, b) (1) For each value of h, this corresponds to calculating the row sum across the table of the joint PMF. Similarly, the marginal PMF of B is PB (b) = 1 h=−1 PH,B (h, b) (2) For each value of b, this corresponds to the column sum down the table of the joint PMF. The easiest way to calculate these marginal PMFs is to simply sum each row and column: PH,B (h, b) b = 0 b = 2 b = 4 PH (h) h = −1 0 0.4 0.2 0.6 h = 0 0.1 0 0.1 0.2 h = 1 0.1 0.1 0 0.2 PB (b) 0.2 0.5 0.3 (3) Quiz 4.4 To find the constant c, we apply ∞ −∞ ∞ −∞ fX,Y (x, y) dx dy = 1. Specifically, ∞ −∞ ∞ −∞ fX,Y (x, y) dx dy = 2 0 1 0 cxy dx dy (1) = c 2 0 y x2 /2 1 0 dy (2) = (c/2) 2 0 y dy = (c/4)y2 2 0 = c (3) Thus c = 1. To calculate P[A], we write P [A] = A fX,Y (x, y) dx dy (4) To integrate over A, we convert to polar coordinates using the substitutions x = r cos θ, y = r sin θ and dx dy = r dr dθ, yielding Y X 1 1 2 A P [A] = π/2 0 1 0 r2 sin θ cos θ r dr dθ (5) = 1 0 r3 dr π/2 0 sin θ cos θ dθ (6) = r4 /4 1 0 ⎛ ⎝ sin2 θ 2 π/2 0 ⎞ ⎠ = 1/8 (7) 22
- 23. Quiz 4.5 By Theorem 4.8, the marginal PDF of X is fX (x) = ∞ −∞ fX,Y (x, y) dy (1) For x < 0 or x > 1, fX (x) = 0. For 0 ≤ x ≤ 1, fX (x) = 6 5 1 0 (x + y2 ) dy = 6 5 xy + y3 /3 y=1 y=0 = 6 5 (x + 1/3) = 6x + 2 5 (2) The complete expression for the PDf of X is fX (x) = (6x + 2)/5 0 ≤ x ≤ 1 0 otherwise (3) By the same method we obtain the marginal PDF for Y. For 0 ≤ y ≤ 1, fY (y) = ∞ −∞ fX,Y (x, y) dy (4) = 6 5 1 0 (x + y2 ) dx = 6 5 x2 /2 + xy2 x=1 x=0 = 6 5 (1/2 + y2 ) = 3 + 6y2 5 (5) Since fY (y) = 0 for y < 0 or y > 1, the complete expression for the PDF of Y is fY (y) = (3 + 6y2)/5 0 ≤ y ≤ 1 0 otherwise (6) Quiz 4.6 (A) The time required for the transfer is T = L/B. For each pair of values of L and B, we can calculate the time T needed for the transfer. We can write these down on the table for the joint PMF of L and B as follows: PL,B(l, b) b = 14, 400 b = 21, 600 b = 28, 800 l = 518, 400 0.20 (T =36) 0.10 (T =24) 0.05 (T =18) l = 2, 592, 000 0.05 (T =180) 0.10 (T =120) 0.20 (T =90) l = 7, 776, 000 0.00 (T =540) 0.10 (T =360) 0.20 (T =270) From the table, writing down the PMF of T is straightforward. PT (t) = ⎧ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩ 0.05 t = 18 0.1 t = 24 0.2 t = 36, 90 0.1 t = 120 0.05 t = 180 0.2 t = 270 0.1 t = 360 0 otherwise (1) 23
- 24. (B) First, we observe that since 0 ≤ X ≤ 1 and 0 ≤ Y ≤ 1, W = XY satisfies 0 ≤ W ≤ 1. Thus fW (0) = 0 and fW (1) = 1. For 0 < w < 1, we calculate the CDF FW (w) = P[W ≤ w]. As shown below, integrating over the region W ≤ w is fairly complex. The calculus is simpler if we integrate over the region XY > w. Specifically, Y X 1 1 XY > w w w XY = w FW (w) = 1 − P [XY > w] (2) = 1 − 1 w 1 w/x dy dx (3) = 1 − 1 w (1 − w/x) dx (4) = 1 − x − w ln x|x=1 x=w (5) = 1 − (1 − w + w ln w) = w − w ln w (6) The complete expression for the CDF is FW (w) = ⎧ ⎨ ⎩ 0 w < 0 w − w ln w 0 ≤ w ≤ 1 1 w > 1 (7) By taking the derivative of the CDF, we find the PDF is fW (w) = d FW (w) dw = ⎧ ⎨ ⎩ 0 w < 0 − ln w 0 ≤ w ≤ 1 0 w > 1 (8) Quiz 4.7 (A) It is helpful to first make a table that includes the marginal PMFs. PL,T (l, t) t = 40 t = 60 PL(l) l = 1 0.15 0.1 0.25 l = 2 0.3 0.2 0.5 l = 3 0.15 0.1 0.25 PT (t) 0.6 0.4 (1) The expected value of L is E [L] = 1(0.25) + 2(0.5) + 3(0.25) = 2. (1) Since the second moment of L is E L2 = 12 (0.25) + 22 (0.5) + 32 (0.25) = 4.5, (2) the variance of L is Var [L] = E L2 − (E [L])2 = 0.5. (3) 24
- 25. (2) The expected value of T is E [T ] = 40(0.6) + 60(0.4) = 48. (4) The second moment of T is E T 2 = 402 (0.6) + 602 (0.4) = 2400. (5) Thus Var[T ] = E T 2 − (E [T ])2 = 2400 − 482 = 96. (6) (3) The correlation is E [LT ] = t=40,60 3 l=1 lt PLT (lt) (7) = 1(40)(0.15) + 2(40)(0.3) + 3(40)(0.15) (8) + 1(60)(0.1) + 2(60)(0.2) + 3(60)(0.1) (9) = 96 (10) (4) From Theorem 4.16(a), the covariance of L and T is Cov [L, T ] = E [LT ] − E [L] E [T ] = 96 − 2(48) = 0 (11) (5) Since Cov[L, T ] = 0, the correlation coefficient is ρL,T = 0. (B) As in the discrete case, the calculations become easier if we first calculate the marginal PDFs fX (x) and fY (y). For 0 ≤ x ≤ 1, fX (x) = ∞ −∞ fX,Y (x, y) dy = 2 0 xy dy = 1 2 xy2 y=2 y=0 = 2x (12) Similarly, for 0 ≤ y ≤ 2, fY (y) = ∞ −∞ fX,Y (x, y) dx = 2 0 xy dx = 1 2 x2 y x=1 x=0 = y 2 (13) The complete expressions for the marginal PDFs are fX (x) = 2x 0 ≤ x ≤ 1 0 otherwise fY (y) = y/2 0 ≤ y ≤ 2 0 otherwise (14) From the marginal PDFs, it is straightforward to calculate the various expectations. 25
- 26. (1) The first and second moments of X are E [X] = ∞ −∞ x fX (x) dx = 1 0 2x2 dx = 2 3 (15) E X2 = ∞ −∞ x2 fX (x) dx = 1 0 2x3 dx = 1 2 (16) (17) The variance of X is Var[X] = E[X2] − (E[X])2 = 1/18. (2) The first and second moments of Y are E [Y] = ∞ −∞ y fY (y) dy = 2 0 1 2 y2 dy = 4 3 (18) E Y2 = ∞ −∞ y2 fY (y) dy = 2 0 1 2 y3 dy = 2 (19) The variance of Y is Var[Y] = E[Y2] − (E[Y])2 = 2 − 16/9 = 2/9. (3) The correlation of X and Y is E [XY] = ∞ −∞ ∞ −∞ xy fX,Y (x, y) dx, dy (20) = 1 0 2 0 x2 y2 dx, dy = x3 3 1 0 y3 3 2 0 = 8 9 (21) (4) The covariance of X and Y is Cov [X, Y] = E [XY] − E [X] E [Y] = 8 9 − 2 3 4 3 = 0. (22) (5) Since Cov[X, Y] = 0, the correlation coefficient is ρX,Y = 0. Quiz 4.8 (A) Since the event V > 80 occurs only for the pairs (L, T ) = (2, 60), (L, T ) = (3, 40) and (L, T ) = (3, 60), P [A] = P [V > 80] = PL,T (2, 60) + PL,T (3, 40) + PL,T (3, 60) = 0.45 (1) By Definition 4.9, PL,T|A (l, t) = PL,T (l,t) P[A] lt > 80 0 otherwise (2) 26
- 27. We can represent this conditional PMF in the following table: PL,T |A(l, t) t = 40 t = 60 l = 1 0 0 l = 2 0 4/9 l = 3 1/3 2/9 The conditional expectation of V can be found from the conditional PMF. E [V |A] = l t lt PL,T |A (l, t) (3) = (2 · 60) 4 9 + (3 · 40) 1 3 + (3 · 60) 2 9 = 133 1 3 (4) For the conditional variance Var[V |A], we first find the conditional second moment E V 2 |A = l t (lt)2 PL,T |A (l, t) (5) = (2 · 60)2 4 9 + (3 · 40)2 1 3 + (3 · 60)2 2 9 = 18, 400 (6) It follows that Var [V |A] = E V 2 |A − (E [V |A])2 = 622 2 9 (7) (B) For continuous random variables X and Y, we first calculate the probability of the conditioning event. P [B] = B fX,Y (x, y) dx dy = 60 40 3 80/y xy 4000 dx dy (8) = 60 40 y 4000 x2 2 3 80/y dy (9) = 60 40 y 4000 9 2 − 3200 y2 dy (10) = 9 8 − 4 5 ln 3 2 ≈ 0.801 (11) The conditional PDF of X and Y is fX,Y|B (x, y) = fX,Y (x, y) /P [B] (x, y) ∈ B 0 otherwise (12) = K xy 40 ≤ y ≤ 60, 80/y ≤ x ≤ 3 0 otherwise (13) 27
- 28. where K = (4000P[B])−1. The conditional expectation of W given event B is E [W|B] = ∞ −∞ ∞ −∞ xy fX,Y|B (x, y) dx dy (14) = 60 40 3 80/y K x2 y2 dx dy (15) = (K/3) 60 40 y2 x3 x=3 x=80/y dy (16) = (K/3) 60 40 27y2 − 803 /y dy (17) = (K/3) 9y3 − 803 ln y 60 40 ≈ 120.78 (18) The conditional second moment of K given B is E W2 |B = ∞ −∞ ∞ −∞ (xy)2 fX,Y|B (x, y) dx dy (19) = 60 40 3 80/y K x3 y3 dx dy (20) = (K/4) 60 40 y3 x4 x=3 x=80/y dy (21) = (K/4) 60 40 81y3 − 804 /y dy (22) = (K/4) (81/4)y4 − 804 ln y 60 40 ≈ 16, 116.10 (23) It follows that the conditional variance of W given B is Var [W|B] = E W2 |B − (E [W|B])2 ≈ 1528.30 (24) Quiz 4.9 (A) (1) The joint PMF of A and B can be found from the marginal and conditional PMFs via PA,B(a, b) = PB|A(b|a)PA(a). Incorporating the information from the given conditional PMFs can be confusing, however. Consequently, we can note that A has range SA = {0, 2} and B has range SB = {0, 1}. A table of the joint PMF will include all four possible combinations of A and B. The general form of the table is PA,B(a, b) b = 0 b = 1 a = 0 PB|A(0|0)PA(0) PB|A(1|0)PA(0) a = 2 PB|A(0|2)PA(2) PB|A(1|2)PA(2) 28
- 29. Substituting values from PB|A(b|a) and PA(a), we have PA,B(a, b) b = 0 b = 1 a = 0 (0.8)(0.4) (0.2)(0.4) a = 2 (0.5)(0.6) (0.5)(0.6) or PA,B(a, b) b = 0 b = 1 a = 0 0.32 0.08 a = 2 0.3 0.3 (2) Given the conditional PMF PB|A(b|2), it is easy to calculate the conditional expectation E [B|A = 2] = 1 b=0 bPB|A (b|2) = (0)(0.5) + (1)(0.5) = 0.5 (1) (3) From the joint PMF PA,B(a, b), we can calculate the the conditional PMF PA|B (a|0) = PA,B (a, 0) PB (0) = ⎧ ⎨ ⎩ 0.32/0.62 a = 0 0.3/0.62 a = 2 0 otherwise (2) = ⎧ ⎨ ⎩ 16/31 a = 0 15/31 a = 2 0 otherwise (3) (4) We can calculate the conditional variance Var[A|B = 0] using the conditional PMF PA|B(a|0). First we calculate the conditional expected value E [A|B = 0] = a aPA|B (a|0) = 0(16/31) + 2(15/31) = 30/31 (4) The conditional second moment is E A2 |B = 0 = a a2 PA|B (a|0) = 02 (16/31) + 22 (15/31) = 60/31 (5) The conditional variance is then Var[A|B = 0] = E A2 |B = 0 − (E [A|B = 0])2 = 960 961 (6) (B) (1) The joint PDF of X and Y is fX,Y (x, y) = fY|X (y|x) fX (x) = 6y 0 ≤ y ≤ x, 0 ≤ x ≤ 1 0 otherwise (7) (2) From the given conditional PDF fY|X (y|x), fY|X (y|1/2) = 8y 0 ≤ y ≤ 1/2 0 otherwise (8) 29
- 30. (3) The conditional PDF of Y given X = 1/2 is fX|Y (x|1/2) = fX,Y (x, 1/2)/fY (1/2). To find fY (1/2), we integrate the joint PDF. fY (1/2) = ∞ −∞ fX,1/2 ( ) dx = 1 1/2 6(1/2) dx = 3/2 (9) Thus, for 1/2 ≤ x ≤ 1, fX|Y (x|1/2) = fX,Y (x, 1/2) fY (1/2) = 6(1/2) 3/2 = 2 (10) (4) From the pervious part, we see that given Y = 1/2, the conditional PDF of X is uniform (1/2, 1). Thus, by the definition of the uniform (a, b) PDF, Var [X|Y = 1/2] = (1 − 1/2)2 12 = 1 48 (11) Quiz 4.10 (A) (1) For random variables X and Y from Example 4.1, we observe that PY (1) = 0.09 and PX (0) = 0.01. However, PX,Y (0, 1) = 0 = PX (0) PY (1) (1) Since we have found a pair x, y such that PX,Y (x, y) = PX (x)PY (y), we can conclude that X and Y are dependent. Note that whenever PX,Y (x, y) = 0, independence requires that either PX (x) = 0 or PY (y) = 0. (2) For random variables Q and G from Quiz 4.2, it is not obvious whether they are independent. Unlike X and Y in part (a), there are no obvious pairs q, g that fail the independence requirement. In this case, we calculate the marginal PMFs from the table of the joint PMF PQ,G(q, g) in Quiz 4.2. PQ,G(q, g) g = 0 g = 1 g = 2 g = 3 PQ(q) q = 0 0.06 0.18 0.24 0.12 0.60 q = 1 0.04 0.12 0.16 0.08 0.40 PG(g) 0.10 0.30 0.40 0.20 Careful study of the table will verify that PQ,G(q, g) = PQ(q)PG(g) for every pair q, g. Hence Q and G are independent. (B) (1) Since X1 and X2 are independent, fX1,X2 (x1, x2) = fX1 (x1) fX2 (x2) (2) = (1 − x1/2)(1 − x2/2) 0 ≤ x1 ≤ 2, 0 ≤ x2 ≤ 2 0 otherwise (3) 30
- 31. (2) Let FX (x) denote the CDF of both X1 and X2. The CDF of Z = max(X1, X2) is found by observing that Z ≤ z iff X1 ≤ z and X2 ≤ z. That is, P [Z ≤ z] = P [X1 ≤ z, X2 ≤ z] (4) = P [X1 ≤ z] P [X2 ≤ z] = [FX (z)]2 (5) To complete the problem, we need to find the CDF of each Xi . From the PDF fX (x), the CDF is FX (x) = x −∞ fX (y) dy = ⎧ ⎨ ⎩ 0 x < 0 x − x2/4 0 ≤ x ≤ 2 1 x > 2 (6) Thus for 0 ≤ z ≤ 2, FZ (z) = (z − z2 /4)2 (7) The complete expression for the CDF of Z is FZ (z) = ⎧ ⎨ ⎩ 0 z < 0 (z − z2/4)2 0 ≤ z ≤ 2 1 z > 1 (8) Quiz 4.11 This problem just requires identifying the various terms in Definition 4.17 and Theo- rem 4.29. Specifically, from the problem statement, we know that ρ = 1/2, µ1 = µX = 0, µ2 = µY = 0, (1) and that σ1 = σX = 1, σ2 = σY = 1. (2) (1) Applying these facts to Definition 4.17, we have fX,Y (x, y) = 1 √ 3π2 e−2(x2−xy+y2)/3 . (3) (2) By Theorem 4.30, the conditional expected value and standard deviation of X given Y = y are E [X|Y = y] = y/2 ˜σX = σ2 1 (1 − ρ2 ) = 3/4. (4) When Y = y = 2, we see that E[X|Y = 2] = 1 and Var[X|Y = 2] = 3/4. The conditional PDF of X given Y = 2 is simply the Gaussian PDF fX|Y (x|2) = 1 √ 3π/2 e−2(x−1)2/3 . (5) 31
- 32. Quiz 4.12 One straightforward method is to follow the approach of Example 4.28. Instead, we use an alternate approach. First we observe that X has the discrete uniform (1, 4) PMF. Also, given X = x, Y has a discrete uniform (1, x) PMF. That is, PX (x) = 1/4 x = 1, 2, 3, 4, 0 otherwise, PY|X (y|x) = 1/x y = 1, . . . , x 0 otherwise (1) Given X = x, and an independent uniform (0, 1) random variable U, we can generate a sample value of Y with a discrete uniform (1, x) PMF via Y = xU . This observation prompts the following program: function xy=dtrianglerv(m) sx=[1;2;3;4]; px=0.25*ones(4,1); x=finiterv(sx,px,m); y=ceil(x.*rand(m,1)); xy=[x’;y’]; 32
- 33. Quiz Solutions – Chapter 5 Quiz 5.1 We find P[C] by integrating the joint PDF over the region of interest. Specifically, P [C] = 1/2 0 dy2 y2 0 dy1 1/2 0 dy4 y4 0 4dy3 (1) = 4 1/2 0 y2 dy2 1/2 0 y4 dy4 = 1/4. (2) Quiz 5.2 By definition of A, Y1 = X1, Y2 = X2 − X1 and Y3 = X3 − X2. Since 0 < X1 < X2 < X3, each Yi must be a strictly positive integer. Thus, for y1, y2, y3 ∈ {1, 2, . . .}, PY (y) = P [Y1 = y1, Y2 = y2, Y3 = y3] (1) = P [X1 = y1, X2 − X1 = y2, X3 − X2 = y3] (2) = P [X1 = y1, X2 = y2 + y1, X3 = y3 + y2 + y1] (3) = (1 − p)3 py1+y2+y3 (4) By defining the vector a = 1 1 1 , the complete expression for the joint PMF of Y is PY (y) = (1 − p)pa y y1, y2, y3 ∈ {1, 2, . . .} 0 otherwise (5) Quiz 5.3 First we note that each marginal PDF is nonzero only if any subset of the xi obeys the ordering contraints 0 ≤ x1 ≤ x2 ≤ x3 ≤ 1. Within these constraints, we have fX1,X2 (x1, x2) = ∞ −∞ fX (x) dx3 = 1 x2 6 dx3 = 6(1 − x2), (1) fX2,X3 (x2, x3) = ∞ −∞ fX (x) dx1 = x2 0 6 dx1 = 6x2, (2) fX1,X3 (x1, x3) = ∞ −∞ fX (x) dx2 = x3 x1 6 dx2 = 6(x3 − x1). (3) In particular, we must keep in mind that fX1,X2(x1, x2) = 0 unless 0 ≤ x1 ≤ x2 ≤ 1, fX2,X3(x2, x3) = 0 unless 0 ≤ x2 ≤ x3 ≤ 1, and that fX1,X3(x1, x3) = 0 unless 0 ≤ x1 ≤ 33
- 34. x3 ≤ 1. The complete expressions are fX1,X2 (x1, x2) = 6(1 − x2) 0 ≤ x1 ≤ x2 ≤ 1 0 otherwise (4) fX2,X3 (x2, x3) = 6x2 0 ≤ x2 ≤ x3 ≤ 1 0 otherwise (5) fX1,X3 (x1, x3) = 6(x3 − x1) 0 ≤ x1 ≤ x3 ≤ 1 0 otherwise (6) Now we can find the marginal PDFs. When 0 ≤ xi ≤ 1 for each xi , fX1 (x1) = ∞ −∞ fX1,X2 (x1, x2) dx2 = 1 x1 6(1 − x2) dx2 = 3(1 − x1)2 (7) fX2 (x2) = ∞ −∞ fX2,X3 (x2, x3) dx3 = 1 x2 6x2 dx3 = 6x2(1 − x2) (8) fX3 (x3) = ∞ −∞ fX2,X3 (x2, x3) dx2 = x3 0 6x2 dx2 = 3x2 3 (9) The complete expressions are fX1 (x1) = 3(1 − x1)2 0 ≤ x1 ≤ 1 0 otherwise (10) fX2 (x2) = 6x2(1 − x2) 0 ≤ x2 ≤ 1 0 otherwise (11) fX3 (x3) = 3x2 3 0 ≤ x3 ≤ 1 0 otherwise (12) Quiz 5.4 In the PDF fY(y), the components have dependencies as a result of the ordering con- straints Y1 ≤ Y2 and Y3 ≤ Y4. We can separate these constraints by creating the vectors V = Y1 Y2 , W = Y3 Y4 . (1) The joint PDF of V and W is fV,W (v, w) = 4 0 ≤ v1 ≤ v2 ≤ 1, 0 ≤ w1 ≤ w2 ≤ 1 0 otherwise (2) 34
- 35. We must verify that V and W are independent. For 0 ≤ v1 ≤ v2 ≤ 1, fV (v) = fV,W (v, w) dw1 dw2 (3) = 1 0 1 w1 4 dw2 dw1 (4) = 1 0 4(1 − w1) dw1 = 2 (5) Similarly, for 0 ≤ w1 ≤ w2 ≤ 1, fW (w) = fV,W (v, w) dv1 dv2 (6) = 1 0 1 v1 4 dv2 dv1 = 2 (7) It follows that V and W have PDFs fV (v) = 2 0 ≤ v1 ≤ v2 ≤ 1 0 otherwise , fW (w) = 2 0 ≤ w1 ≤ w2 ≤ 1 0 otherwise (8) It is easy to verify that fV,W(v, w) = fV(v) fW(w), confirming that V and W are indepen- dent vectors. Quiz 5.5 (A) Referring to Theorem 1.19, each test is a subexperiment with three possible out- comes: L, A and R. In five trials, the vector X = X1 X2 X3 indicating the number of outcomes of each subexperiment has the multinomial PMF PX (x) = ⎧ ⎨ ⎩ 5 x1,x2,x3 (0.3)x1(0.6)x2(0.1)x3 x1 + x2 + x3 = 5; x1, x2, x3 ∈ {0, 1, . . . , 5} 0 otherwise (1) We can find the marginal PMF for each Xi from the joint PMF PX(x); however it is simpler to just start from first principles and observe that X1 is the number of occurrences of L in five independent tests. If we view each test as a trial with success probability P[L] = 0.3, we see that X1 is a binomial (n, p) = (5, 0.3) random variable. Similarly, X2 is a binomial (5, 0.6) random variable and X3 is a binomial (5, 0.1) random variable. That is, for p1 = 0.3, p2 = 0.6 and p3 = 0.1, PXi (x) = 5 x px i (1 − pi )5−x x = 0, 1, . . . , 5 0 otherwise (2) 35
- 36. From the marginal PMFs, we see that X1, X2 and X3 are not independent. Hence, we must use Theorem 5.6 to find the PMF of W. In particular, since X1 + X2 + X3 = 5 and since each Xi is non-negative, PW (0) = PW (1) = 0. Furthermore, PW (2) = PX (1, 2, 2) + PX (2, 1, 2) + PX (2, 2, 1) (3) = 5![0.3(0.6)2(0.1)2 + 0.32(0.6)(0.1)2 + 0.32(0.6)2(0.1)] 2!2!1! (4) = 0.1458 (5) In addition, for w = 3, w = 4, and w = 5, the event W = w occurs if and only if one of the mutually exclusive events X1 = w, X2 = w, or X3 = w occurs. Thus, PW (3) = PX1 (3) + PX2 (3) + PX3 (3) = 0.486 (6) PW (4) = PX1 (4) + PX2 (4) + PX3 (4) = 0.288 (7) PW (5) = PX1 (5) + PX2 (5) + PX3 (5) = 0.0802 (8) (B) Since each Yi = 2Xi + 4, we can apply Theorem 5.10 to write fY (y) = 1 23 fX y1 − 4 2 , y2 − 4 2 , y3 − 4 2 (9) = (1/8)e−(y3−4)/2 4 ≤ y1 ≤ y2 ≤ y3 0 otherwise (10) Note that for other matrices A, the constraints on y resulting from the constraints 0 ≤ X1 ≤ X2 ≤ X3 can be much more complicated. Quiz 5.6 We start by finding the components E[Xi ] = ∞ −∞ x fXi (x) dx of µX . To do so, we use the marginal PDFs fXi (x) found in Quiz 5.3: E [X1] = 1 0 3x(1 − x)2 dx = 1/4, (1) E [X2] = 1 0 6x2 (1 − x) dx = 1/2, (2) E [X3] = 1 0 3x3 dx = 3/4. (3) To find the correlation matrix RX , we need to find E[Xi X j ] for all i and j. We start with 36
- 37. the second moments: E X2 1 = 1 0 3x2 (1 − x)2 dx = 1/10. (4) E X2 2 = 1 0 6x3 (1 − x) dx = 3/10. (5) E X2 3 = 1 0 3x4 dx = 3/5. (6) Using marginal PDFs from Quiz 5.3, the cross terms are E [X1 X2] = ∞ −∞ ∞ −∞ x1x2 fX1,X2 (x1, x2) , dx1 dx2 (7) = 1 0 1 x1 6x1x2(1 − x2) dx2 dx1 (8) = 1 0 [x1 − 3x3 1 + 2x4 1] dx1 = 3/20. (9) E [X2 X3] = 1 0 1 x2 6x2 2 x3 dx3 dx2 (10) = 1 0 [3x2 2 − 3x4 2] dx2 = 2/5 (11) E [X1 X3] = 1 0 1 x1 6x1x3(x3 − x1) dx3 dx1. (12) = 1 0 (2x1x3 3 − 3x2 1 x2 3) x3=1 x3=x1 dx1 (13) = 1 0 [2x1 − 3x2 1 + x4 1] dx1 = 1/5. (14) Summarizing the results, X has correlation matrix RX = ⎡ ⎣ 1/10 3/20 1/5 3/20 3/10 2/5 1/5 2/5 3/5 ⎤ ⎦ . (15) Vector X has covariance matrix CX = RX − E [X] E [X] (16) = ⎡ ⎣ 1/10 3/20 1/5 3/20 3/10 2/5 1/5 2/5 3/5 ⎤ ⎦ − ⎡ ⎣ 1/4 1/2 3/4 ⎤ ⎦ 1/4 1/2 3/4 (17) = ⎡ ⎣ 1/10 3/20 1/5 3/20 3/10 2/5 1/5 2/5 3/5 ⎤ ⎦ − ⎡ ⎣ 1/16 1/8 3/16 1/8 1/4 3/8 3/16 3/8 9/16 ⎤ ⎦ = 1 80 ⎡ ⎣ 3 2 1 2 4 2 1 2 3 ⎤ ⎦ . (18) 37
- 38. This problem shows that even for fairly simple joint PDFs, computing the covariance matrix by calculus can be a time consuming task. Quiz 5.7 We observe that X = AZ + b where A = 2 1 1 −1 , b = 2 0 . (1) It follows from Theorem 5.18 that µX = b and that CX = AA = 2 1 1 −1 2 1 1 −1 = 5 1 1 2 . (2) Quiz 5.8 First, we observe that Y = AT where A = 1/31 1/31 · · · 1/31 . Since T is a Gaussian random vector, Theorem 5.16 tells us that Y is a 1 dimensional Gaussian vector, i.e., just a Gaussian random variable. The expected value of Y is µY = µT = 80. The covariance matrix of Y is 1 × 1 and is just equal to Var[Y]. Thus, by Theorem 5.16, Var[Y] = ACT A . function p=julytemps(T); [D1 D2]=ndgrid((1:31),(1:31)); CT=36./(1+abs(D1-D2)); A=ones(31,1)/31.0; CY=(A’)*CT*A; p=phi((T-80)/sqrt(CY)); In julytemps.m, the first two lines gen- erate the 31 × 31 covariance matrix CT, or CT . Next we calculate Var[Y]. The final step is to use the (·) function to calculate P[Y < T ]. Here is the output of julytemps.m: >> julytemps([70 75 80 85 90 95]) ans = 0.0000 0.0221 0.5000 0.9779 1.0000 1.0000 Note that P[T ≤ 70] is not actually zero and that P[T ≤ 90] is not actually 1.0000. Its just that the MATLAB’s short format output, invoked with the command format short, rounds off those probabilities. Here is the long format output: >> format long >> julytemps([70 75 80 85 90 95]) ans = Columns 1 through 4 0.00002844263128 0.02207383067604 0.50000000000000 0.97792616932396 Columns 5 through 6 0.99997155736872 0.99999999922010 38
- 39. The ndgrid function is a useful to way calculate many covariance matrices. However, in this problem, CX has a special structure; the i, jth element is CT(i, j) = c|i− j| = 36 1 + |i − j| . (1) If we write out the elements of the covariance matrix, we see that CT = ⎡ ⎢ ⎢ ⎢ ⎣ c0 c1 · · · c30 c1 c0 ... ... ... ... ... c1 c30 · · · c1 c0 ⎤ ⎥ ⎥ ⎥ ⎦ . (2) This covariance matrix is known as a symmetric Toeplitz matrix. We will see in Chap- ters 9 and 11 that Toeplitz covariance matrices are quite common. In fact, MATLAB has a toeplitz function for generating them. The function julytemps2 use the toeplitz to generate the correlation matrix CT. function p=julytemps2(T); c=36./(1+abs(0:30)); CT=toeplitz(c); A=ones(31,1)/31.0; CY=(A’)*CT*A; p=phi((T-80)/sqrt(CY)); 39
- 40. Quiz Solutions – Chapter 6 Quiz 6.1 Let K1, . . . , Kn denote a sequence of iid random variables each with PMF PK (k) = 1/4 k = 1, . . . , 4 0 otherwise (1) We can write Wn in the form of Wn = K1 + · · · + Kn. First, we note that the first two moments of Ki are E [Ki ] = (1 + 2 + 3 + 4)/4 = 2.5 (2) E K2 i = (12 + 22 + 32 + 42 )/4 = 7.5 (3) Thus the variance of Ki is Var[Ki ] = E K2 i − (E [Ki ])2 = 7.5 − (2.5)2 = 1.25 (4) Since E[Ki ] = 2.5, the expected value of Wn is E [Wn] = E [K1] + · · · + E [Kn] = nE [Ki ] = 2.5n (5) Since the rolls are independent, the random variables K1, . . . , Kn are independent. Hence, by Theorem 6.3, the variance of the sum equals the sum of the variances. That is, Var[Wn] = Var[K1] + · · · + Var[Kn] = 1.25n (6) Quiz 6.2 Random variables X and Y have PDFs fX (x) = 3e−3x x ≥ 0 0 otherwise fY (y) = 2e−2y y ≥ 0 0 otherwise (1) Since X and Y are nonnegative, W = X + Y is nonnegative. By Theorem 6.5, the PDF of W = X + Y is fW (w) = ∞ −∞ fX (w − y) fY (y) dy = 6 w 0 e−3(w−y) e−2y dy (2) Fortunately, this integral is easy to evaluate. For w > 0, fW (w) = e−3w ey w 0 = 6 e−2w − e−3w (3) Since fW (w) = 0 for w < 0, a conmplete expression for the PDF of W is fW (w) = 6e−2w 1 − e−w w ≥ 0, 0 otherwise. (4) 40
- 41. Quiz 6.3 The MGF of K is φK (s) = E esK == 4 k=0 (0.2)esk = 0.2 1 + es + e2s + e3s + e4s (1) We find the moments by taking derivatives. The first derivative of φK (s) is dφK (s) ds = 0.2(es + 2e2s + 3e3s + 4e4s ) (2) Evaluating the derivative at s = 0 yields E [K] = dφK (s) ds s=0 = 0.2(1 + 2 + 3 + 4) = 2 (3) To find higher-order moments, we continue to take derivatives: E K2 = d2φK (s) ds2 s=0 = 0.2(es + 4e2s + 9e3s + 16e4s ) s=0 = 6 (4) E K3 = d3φK (s) ds3 s=0 = 0.2(es + 8e2s + 27e3s + 64e4s ) s=0 = 20 (5) E K4 = d4φK (s) ds4 s=0 = 0.2(es + 16e2s + 81e3s + 256e4s ) s=0 = 70.8 (6) (7) Quiz 6.4 (A) Each Ki has MGF φK (s) = E esKi = es + e2s + · · · + ens n = es(1 − ens) n(1 − es) (1) Since the sequence of Ki is independent, Theorem 6.8 says the MGF of J is φJ (s) = (φK (s))m = ems(1 − ens)m nm(1 − es)m (2) (B) Since the set of α j X j are independent Gaussian random variables, Theorem 6.10 says that W is a Gaussian random variable. Thus to find the PDF of W, we need only find the expected value and variance. Since the expectation of the sum equals the sum of the expectations: E [W] = αE [X1] + α2 E [X2] + · · · + αn E [Xn] = 0 (3) 41
- 42. Since the α j X j are independent, the variance of the sum equals the sum of the vari- ances: Var[W] = α2 Var[X1] + α4 Var[X2] + · · · + α2n Var[Xn] (4) = α2 + 2(α2 )2 + 3(α2 )3 + · · · + n(α2 )n (5) Defining q = α2, we can use Math Fact B.6 to write Var[W] = α2 − α2n+2[1 + n(1 − α2)] (1 − α2)2 (6) With E[W] = 0 and σ2 W = Var[W], we can write the PDF of W as fW (w) = 1 2πσ2 W e−w2/2σ2 W (7) Quiz 6.5 (1) From Table 6.1, each Xi has MGF φX (s) and random variable N has MGF φN (s) where φX (s) = 1 1 − s , φN (s) = 1 5es 1 − 4 5es . (1) From Theorem 6.12, R has MGF φR(s) = φN (ln φX (s)) = 1 5φX (s) 1 − 4 5φX (s) (2) Substituting the expression for φX (s) yields φR(s) = 1 5 1 5 − s . (3) (2) From Table 6.1, we see that R has the MGF of an exponential (1/5) random variable. The corresponding PDF is fR (r) = (1/5)e−r/5 r ≥ 0 0 otherwise (4) This quiz is an example of the general result that a geometric sum of exponential random variables is an exponential random variable. 42
- 43. Quiz 6.6 (1) The expected access time is E [X] = ∞ −∞ x fX (x) dx = 12 0 x 12 dx = 6 msec (1) (2) The second moment of the access time is E X2 = ∞ −∞ x2 fX (x) dx = 12 0 x2 12 dx = 48 (2) The variance of the access time is Var[X] = E[X2] − (E[X])2 = 48 − 36 = 12. (3) Using Xi to denote the access time of block i, we can write A = X1 + X2 + · · · + X12 (3) Since the expectation of the sum equals the sum of the expectations, E [A] = E [X1] + · · · + E [X12] = 12E [X] = 72 msec (4) (4) Since the Xi are independent, Var[A] = Var[X1] + · · · + Var[X12] = 12 Var[X] = 144 (5) Hence, the standard deviation of A is σA = 12 (5) To use the central limit theorem, we write P [A > 75] = 1 − P [A ≤ 75] (6) = 1 − P A − E [A] σA ≤ 75 − E [A] σA (7) ≈ 1 − 75 − 72 12 (8) = 1 − 0.5987 = 0.4013 (9) Note that we used Table 3.1 to look up (0.25). (6) Once again, we use the central limit theorem and Table 3.1 to estimate P [A < 48] = P A − E [A] σA < 48 − E [A] σA (10) ≈ 48 − 72 12 (11) = 1 − (2) = 1 − 0.9773 = 0.0227 (12) 43
- 44. Quiz 6.7 Random variable Kn has a binomial distribution for n trials and success probability P[V ] = 3/4. (1) The expected number of voice calls out of 48 calls is E[K48] = 48P[V ] = 36. (2) The variance of K48 is Var[K48] = 48P [V ] (1 − P [V ]) = 48(3/4)(1/4) = 9 (1) Thus K48 has standard deviation σK48 = 3. (3) Using the ordinary central limit theorem and Table 3.1 yields P [30 ≤ K48 ≤ 42] ≈ 42 − 36 3 − 30 − 36 3 = (2) − (−2) (2) Recalling that (−x) = 1 − (x), we have P [30 ≤ K48 ≤ 42] ≈ 2 (2) − 1 = 0.9545 (3) (4) Since K48 is a discrete random variable, we can use the De Moivre-Laplace approx- imation to estimate P [30 ≤ K48 ≤ 42] ≈ 42 + 0.5 − 36 3 − 30 − 0.5 − 36 3 (4) = 2 (2.16666) − 1 = 0.9687 (5) Quiz 6.8 The train interarrival times X1, X2, X3 are iid exponential (λ) random variables. The arrival time of the third train is W = X1 + X2 + X3. (1) In Theorem 6.11, we found that the sum of three iid exponential (λ) random variables is an Erlang (n = 3, λ) random variable. From Appendix A, we find that W has expected value and variance E [W] = 3/λ = 6 Var[W] = 3/λ2 = 12 (2) (1) By the Central Limit Theorem, P [W > 20] = P W − 6 √ 12 > 20 − 6 √ 12 ≈ Q(7/ √ 3) = 2.66 × 10−5 (3) 44
- 45. (2) To use the Chernoff bound, we note that the MGF of W is φW (s) = λ λ − s 3 = 1 (1 − 2s)3 (4) The Chernoff bound states that P [W > 20] ≤ min s≥0 e−20s φX (s) = min s≥0 e−20s (1 − 2s)3 (5) To minimize h(s) = e−20s/(1 − 2s)3, we set the derivative of h(s) to zero: dh(s) ds = −20(1 − 2s)3e−20s + 6e−20s(1 − 2s)2 (1 − 2s)6 = 0 (6) This implies 20(1 − 2s) = 6 or s = 7/20. Applying s = 7/20 into the Chernoff bound yields P [W > 20] ≤ e−20s (1 − 2s)3 s=7/20 = (10/3)3 e−7 = 0.0338 (7) (3) Theorem 3.11 says that for any w > 0, the CDF of the Erlang (λ, 3) random variable W satisfies FW (w) = 1 − 2 k=0 (λw)ke−λw k! (8) Equivalently, for λ = 1/2 and w = 20, P [W > 20] = 1 − FW (20) (9) = e−10 1 + 10 1! + 102 2! = 61e−10 = 0.0028 (10) Although the Chernoff bound is relatively weak in that it overestimates the proba- bility by roughly a factor of 12, it is a valid bound. By contrast, the Central Limit Theorem approximation grossly underestimates the true probability. Quiz 6.9 One solution to this problem is to follow the approach of Example 6.19: %unifbinom100.m sx=0:100;sy=0:100; px=binomialpmf(100,0.5,sx); py=duniformpmf(0,100,sy); [SX,SY]=ndgrid(sx,sy); [PX,PY]=ndgrid(px,py); SW=SX+SY; PW=PX.*PY; sw=unique(SW); pw=finitepmf(SW,PW,sw); pmfplot(sw,pw,’itw’,’itP_W(w)’); A graph of the PMF PW (w) appears in Figure 2 With some thought, it should be apparent that the finitepmf function is implementing the convolution of the two PMFs. 45
- 46. 0 20 40 60 80 100 120 140 160 180 200 0 0.002 0.004 0.006 0.008 0.01 w P W (w) Figure 2: From Quiz 6.9, the PMF PW (w) of the independent sum of a binomial (100, 0.5) random variable and a discrete uniform (0, 100) random variable. 46
- 47. Quiz Solutions – Chapter 7 Quiz 7.1 An exponential random variable with expected value 1 also has variance 1. By Theo- rem 7.1, Mn(X) has variance Var[Mn(X)] = 1/n. Hence, we need n = 100 samples. Quiz 7.2 The arrival time of the third elevator is W = X1 + X2 + X3. Since each Xi is uniform (0, 30), E [Xi ] = 15, Var [Xi ] = (30 − 0)2 12 = 75. (1) Thus E[W] = 3E[Xi ] = 45, and Var[W] = 3 Var[Xi ] = 225. (1) By the Markov inequality, P [W > 75] ≤ E [W] 75 = 45 75 = 3 5 (2) (2) By the Chebyshev inequality, P [W > 75] = P [W − E [W] > 30] (3) ≤ P [|W − E [W]| > 30] ≤ Var [W] 302 = 225 900 = 1 4 (4) Quiz 7.3 Define the random variable W = (X − µX )2. Observe that V100(X) = M100(W). By Theorem 7.6, the mean square error is E (M100(W) − µW )2 = Var[W] 100 (1) Observe that µX = 0 so that W = X2. Thus, µW = E X2 = 1 −1 x2 fX (x) dx = 1/3 (2) E W2 = E X4 = 1 −1 x4 fX (x) dx = 1/5 (3) Therefore Var[W] = E[W2] − µ2 W = 1/5 − (1/3)2 = 4/45 and the mean square error is 4/4500 = 0.000889. 47
- 48. Quiz 7.4 Assuming the number n of samples is large, we can use a Gaussian approximation for Mn(X). SinceE[X] = p and Var[X] = p(1 − p), we apply Theorem 7.13 which says that the interval estimate Mn(X) − c ≤ p ≤ Mn(X) + c (1) has confidence coefficient 1 − α where α = 2 − 2 c √ n p(1 − p) . (2) We must ensure for every value of p that 1 − α ≥ 0.9 or α ≤ 0.1. Equivalently, we must have c √ n p(1 − p) ≥ 0.95 (3) for every value of p. Since (x) is an increasing function of x, we must satisfy c √ n ≥ 1.65p(1 − p). Since p(1 − p) ≤ 1/4 for all p, we require that c ≥ 1.65 4 √ n = 0.41 √ n . (4) The 0.9 confidence interval estimate of p is Mn(X) − 0.41 √ n ≤ p ≤ Mn(X) + 0.41 √ n . (5) For the 0.99 confidence interval, we have α ≤ 0.01, implying (c √ n/(p(1−p))) ≥ 0.995. This implies c √ n ≥ 2.58p(1 − p). Since p(1 − p) ≤ 1/4 for all p, we require that c ≥ (0.25)(2.58)/ √ n. In this case, the 0.99 confidence interval estimate is Mn(X) − 0.645 √ n ≤ p ≤ Mn(X) + 0.645 √ n . (6) Note that if M100(X) = 0.4, then the 0.99 confidence interval estimate is 0.3355 ≤ p ≤ 0.4645. (7) The interval is wide because the 0.99 confidence is high. Quiz 7.5 Following the approach of bernoullitraces.m, we generate m = 1000 sample paths, each sample path having n = 100 Bernoulli traces. at time k, OK(k) counts the fraction of sample paths that have sample mean within one standard error of p. The pro- gram bernoullisample.m generates graphs the number of traces within one standard error as a function of the time, i.e. the number of trials in each trace. 48
- 49. function OK=bernoullisample(n,m,p); x=reshape(bernoullirv(p,m*n),n,m); nn=(1:n)’*ones(1,m); MN=cumsum(x)./nn; stderr=sqrt(p*(1-p))./sqrt((1:n)’); stderrmat=stderr*ones(1,m); OK=sum(abs(MN-p)<stderrmat,2)/m; plot(1:n,OK,’-s’); The following graph was generated by bernoullisample(100,5000,0.5): 0 10 20 30 40 50 60 70 80 90 100 0.4 0.5 0.6 0.7 0.8 0.9 1 As we would expect, as m gets large, the fraction of traces within one standard error ap- proaches 2 (1) − 1 ≈ 0.68. The unusual sawtooth pattern, though perhaps unexpected, is examined in Problem 7.5.2. 49
- 50. Quiz Solutions – Chapter 8 Quiz 8.1 From the problem statement, each Xi has PDF and CDF fXi (x) = e−x x ≥ 0 0 otherwise FXi (x) = 0 x < 0 1 − e−x x ≥ 0 (1) Hence, the CDF of the maximum of X1, . . . , X15 obeys FX (x) = P [X ≤ x] = P [X1 ≤ x, X2 ≤ x, · · · , X15 ≤ x] = [P [Xi ≤ x]]15 . (2) This implies that for x ≥ 0, FX (x) = FXi (x) 15 = 1 − e−x 15 (3) To design a significance test, we must choose a rejection region for X. A reasonable choice is to reject the hypothesis if X is too small. That is, let R = {X ≤ r}. For a significance level of α = 0.01, we obtain α = P [X ≤ r] = (1 − e−r )15 = 0.01 (4) It is straightforward to show that r = − ln 1 − (0.01)1/15 = 1.33 (5) Hence, if we observe X < 1.33, then we reject the hypothesis. Quiz 8.2 From the problem statement, the conditional PMFs of K are PK|H0 (k) = 104ke−104 k! k = 0, 1, . . . 0 otherwise (1) PK|H1 (k) = 106ke−106 k! k = 0, 1, . . . 0 otherwise (2) Since the two hypotheses are equally likely, the MAP and ML tests are the same. From Theorem 8.6, the ML hypothesis rule is k ∈ A0 if PK|H0 (k) ≥ PK|H1 (k) ; k ∈ A1 otherwise. (3) This rule simplifies to k ∈ A0 if k ≤ k∗ = 106 − 104 ln 100 = 214, 975.7; k ∈ A1 otherwise. (4) Thus if we observe at least 214, 976 photons, then we accept hypothesis H1. 50
- 51. Quiz 8.3 For the QPSK system, a symbol error occurs when si is transmitted but (X1, X2) ∈ Aj for some j = i. For a QPSK system, it is easier to calculate the probability of a correct decision. Given H0, the conditional probability of a correct decision is P [C|H0] = P [X1 > 0, X2 > 0|H0] = P √ E/2 + N1 > 0, √ E/2 + N2 > 0 (1) Because of the symmetry of the signals, P[C|H0] = P[C|Hi ] for all i. This implies the probability of a correct decision is P[C] = P[C|H0]. Since N1 and N2 are iid Gaussian (0, σ) random variables, we have P [C] = P [C|H0] = P √ E/2 + N1 > 0 P √ E/2 + N2 > 0 (2) = P N1 > − √ E/2 2 (3) = 1 − − √ E/2 σ 2 (4) Since (−x) = 1 − (x), we have P[C] = 2( E/2σ2). Equivalently, the probability of error is PERR = 1 − P [C] = 1 − 2 E 2σ2 (5) Quiz 8.4 To generate the ROC, the existing program sqdistor already calculates this miss probability PMISS = P01 and the false alarm probability PFA = P10. The modified pro- gram, sqdistroc.m is essentially the same as sqdistor except the output is a ma- trix FM whose columns are the false alarm and miss probabilities. Next, the program sqdistrocplot.m calls sqdistroc three times to generate a plot that compares the receiver performance for the three requested values of d. Here is the modified code: function FM=sqdistroc(v,d,m,T) %square law distortion recvr %P(error) for m bits tested %transmit v volts or -v volts, %add N volts, N is Gauss(0,1) %add d(v+N)ˆ2 distortion %receive 1 if x>T, otherwise 0 %FM = [P(FA) P(MISS)] x=(v+randn(m,1)); [XX,TT]=ndgrid(x,T(:)); P01=sum((XX+d*(XX.ˆ2)< TT),1)/m; x= -v+randn(m,1); [XX,TT]=ndgrid(x,T(:)); P10=sum((XX+d*(XX.ˆ2)>TT),1)/m; FM=[P10(:) P01(:)]; function FM=sqdistrocplot(v,m,T); FM1=sqdistroc(v,0.1,m,T); FM2=sqdistroc(v,0.2,m,T); FM5=sqdistroc(v,0.3,m,T); FM=[FM1 FM2 FM5]; loglog(FM1(:,1),FM1(:,2),’-k’, ... FM2(:,1),FM2(:,2),’--k’, ... FM5(:,1),FM5(:,2),’:k’); legend(’it d=0.1’,’it d=0.2’,... ’it d=0.3’,3) ylabel(’P_{MISS}’); xlabel(’P_{FA}’); 51
- 52. To see the effect of d, the commands T=-3:0.1:3; sqdistrocplot(3,100000,T); generated the plot shown in Figure 3. 10 −5 10 −4 10 −3 10 −2 10 −1 10 0 10 −5 10 −4 10 −3 10 −2 10 −1 10 0 P MISS P FA d=0.1 d=0.2 d=0.3 T=-3:0.1:3; sqdistrocplot(3,100000,T); Figure 3: The receiver operating curve for the communications system of Quiz 8.4 with squared distortion. 52
- 53. Quiz Solutions – Chapter 9 Quiz 9.1 (1) First, we calculate the marginal PDF for 0 ≤ y ≤ 1: fY (y) = y 0 2(y + x) dx = 2xy + x2 x=y x=0 = 3y2 (1) This implies the conditional PDF of X given Y is fX|Y (x|y) = fX,Y (x, y) fY (y) = 2 3y + 2x 3y2 0 ≤ x ≤ y 0 otherwise (2) (2) The minimum mean square error estimate of X given Y = y is ˆxM(y) = E [X|Y = y] = y 0 2x 3y + 2x2 3y2 dx = 5y/9 (3) Thus the MMSE estimator of X given Y is ˆXM(Y) = 5Y/9. (3) To obtain the conditional PDF fY|X (y|x), we need the marginal PDF fX (x). For 0 ≤ x ≤ 1, fX (x) = 1 x 2(y + x) dy = y2 + 2xy y=1 y=x = 1 + 2x − 3x2 (4) (5) For 0 ≤ x ≤ 1, the conditional PDF of Y given X is fY|X (y|x) = 2(y+x) 1+2x−3x2 x ≤ y ≤ 1 0 otherwise (6) (4) The MMSE estimate of Y given X = x is ˆyM(x) = E [Y|X = x] = 1 x 2y2 + 2xy 1 + 2x − 3x2 dy (7) = 2y3/3 + xy2 1 + 2x − 3x2 y=1 y=x (8) = 2 + 3x − 5x3 3 + 6x − 9x2 (9) 53
- 54. Quiz 9.2 (1) Since the expectation of the sum equals the sum of the expectations, E [R] = E [T ] + E [X] = 0 (1) (2) Since T and X are independent, the variance of the sum R = T + X is Var[R] = Var[T ] + Var[X] = 9 + 3 = 12 (2) (3) Since T and R have expected values E[R] = E[T ] = 0, Cov [T, R] = E [T R] = E [T (T + X)] = E T 2 + E [T X] (3) Since T and X are independent and have zero expected value, E[T X] = E[T ]E[X] = 0 and E[T 2] = Var[T ]. Thus Cov[T, R] = Var[T ] = 9. (4) From Definition 4.8, the correlation coefficient of T and R is ρT,R = Cov [T, R] √ Var[R] Var[T ] = σT σR = √ 3/2 (4) (5) From Theorem 9.4, the optimum linear estimate of T given R is ˆTL(R) = ρT,R σT σR (R − E [R]) + E [T ] (5) Since E[R] = E[T ] = 0 and ρT,R = σT /σR, ˆTL(R) = σ2 T σ2 R R = σ2 T σ2 T + σ2 X R = 3 4 R (6) Hence a∗ = 3/4 and b∗ = 0. (6) By Theorem 9.4, the mean square error of the linear estimate is e∗ L = Var[T ](1 − ρ2 T,R) = 9(1 − 3/4) = 9/4 (7) Quiz 9.3 When R = r, the conditional PDF of X = Y −40−40 log10 r is Gaussian with expected value −40 − 40 log10 r and variance 64. The conditional PDF of X given R is fX|R (x|r) = 1 √ 128π e−(x+40+40 log10 r)2/128 (1) 54
- 55. From the conditional PDF fX|R(x|r), we can use Definition 9.2 to write the ML estimate of R given X = x as ˆrML(x) = arg max r≥0 fX|R (x|r) (2) We observe that fX|R(x|r) is maximized when the exponent (x + 40 + 40 log10 r)2 is minimized. This minimum occurs when the exponent is zero, yielding log10 r = −1 − x/40 (3) or ˆrML(x) = (0.1)10−x/40 m (4) If the result doesn’t look correct, note that a typical figure for the signal strength might be x = −120 dB. This corresponds to a distance estimate of ˆrML(−120) = 100 m. For the MAP estimate, we observe that the joint PDF of X and R is fX,R (x,r) = fX|R (x|r) fR (r) = 1 106 √ 32π re−(x+40+40 log10 r)2/128 (5) From Theorem 9.6, the MAP estimate of R given X = x is the value of r that maximizes fX,R(x,r). That is, ˆrMAP(x) = arg max 0≤r≤1000 fX,R (x,r) (6) Note that we have included the constraint r ≤ 1000 in the maximization to highlight the fact that under our probability model, R ≤ 1000 m. Setting the derivative of fX,R(x,r) with respect to r to zero yields e−(x+40+40 log10 r)2/128 1 − 80 log10 e 128 (x + 40 + 40 log10 r) = 0 (7) Solving for r yields r = 10 1 25 log10 e −1 10−x/40 = (0.1236)10−x/40 (8) This is the MAP estimate of R given X = x as long as r ≤ 1000 m. When x ≤ −156.3 dB, the above estimate will exceed 1000 m, which is not possible in our probability model. Hence, the complete description of the MAP estimate is ˆrMAP(x) = 1000 x < −156.3 (0.1236)10−x/40 x ≥ −156.3 (9) For example, if x = −120dB, then ˆrMAP(−120) = 123.6 m. When the measured signal strength is not too low, the MAP estimate is 23.6% larger than the ML estimate. This re- flects the fact that large values of R are a priori more probable than small values. However, for very low signal strengths, the MAP estimate takes into account that the distance can never exceed 1000 m. 55
- 56. Quiz 9.4 (1) From Theorem 9.4, the LMSE estimate of X2 given Y2 is ˆX2(Y2) = a∗Y2 +b∗ where a∗ = Cov [X2, Y2] Var[Y2] , b∗ = µX2 − a∗ µY2. (1) Because E[X] = E[Y] = 0, Cov [X2, Y2] = E [X2Y2] = E [X2(X2 + W2)] = E X2 2 = 1 (2) Var[Y2] = Var[X2] + Var[W2] = E X2 2 + E W2 2 = 1.1 (3) It follows that a∗ = 1/1.1. Because µX2 = µY2 = 0, it follows that b∗ = 0. Finally, to compute the expected square error, we calculate the correlation coefficient ρX2,Y2 = Cov [X2, Y2] σX2σY2 = 1 √ 1.1 (4) The expected square error is e∗ L = Var[X2](1 − ρ2 X2,Y2 ) = 1 − 1 1.1 = 1 11 = 0.0909 (5) (2) Since Y = X + W and E[X] = E[W] = 0, it follows that E[Y] = 0. Thus we can apply Theorem 9.7. Note that X and W have correlation matrices RX = 1 −0.9 −0.9 1 , RW = 0.1 0 0 0.1 . (6) In terms of Theorem 9.7, n = 2 and we wish to estimate X2 given the observation vector Y = Y1 Y2 . To apply Theorem 9.7, we need to find RY and RYX2. RY = E YY = E (X + W)(X + W ) (7) = E XX + XW + WX + WW . (8) Because X and W are independent, E[XW ] = E[X]E[W ] = 0. Similarly, E[WX ] = 0. This implies RY = E XX + E WW = RX + RW = 1.1 −0.9 −0.9 1.1 . (9) In addition, we need to find RYX2 = E [YX2] = E [Y1 X2] E [Y2 X2] = E [(X1 + W1)X2] E [(X2 + W2)X2] . (10) 56
- 57. Since X and W are independent vectors, E[W1 X2] = E[W1]E[X2] = 0 and E[W2 X2] = 0. Thus RYX2 = E[X1 X2] E X2 2 = −0.9 1 . (11) By Theorem 9.7, ˆa = R−1 Y RYX2 = −0.225 0.725 (12) Therefore, the optimum linear estimator of X2 given Y1 and Y2 is ˆXL = ˆa Y = −0.225Y1 + 0.725Y2. (13) The mean square error is Var [X2] − ˆa RYX2 = Var [X] − a1rY1,X2 − a2rY2,X2 = 0.0725. (14) Quiz 9.5 Since X and W have zero expected value, Y also has zero expected value. Thus, by Theorem 9.7, ˆXL(Y) = ˆa Y where ˆa = R−1 Y RYX . Since X and W are independent, E[WX] = 0 and E[XW ] = 0 . This implies RYX = E [YX] = E [(1X + W)X] = 1E X2 = 1. (1) By the same reasoning, the correlation matrix of Y is RY = E YY = E (1X + W)(1 X + W ) (2) = 11 E X2 + 1E XW + E [WX] 1 + E WW (3) = 11 + RW (4) Note that 11 is a 20 × 20 matrix with every entry equal to 1. Thus, ˆa = R−1 Y RYX = 11 + RW −1 1 (5) and the optimal linear estimator is ˆXL(Y) = 1 11 + RW −1 Y (6) The mean square error is e∗ L = Var[X] − ˆa RYX = 1 − 1 11 + RW −1 1 (7) Now we note that RW has i, jth entry RW(i, j) = c|i− j|−1. The question we must address is what value c minimizes e∗ L. This problem is atypical in that one does not usually get 57
- 58. to choose the correlation structure of the noise. However, we will see that the answer is somewhat instructive. We note that the answer is not obviously apparent from Equation (7). In particular, we observe that Var[Wi ] = RW(i, i) = 1/c. Thus, when c is small, the noises Wi have high variance and we would expect our estimator to be poor. On the other hand, if c is large Wi and Wj are highly correlated and the separate measurements of X are very dependent. This would suggest that large values of c will also result in poor MSE. If this argument is not clear, consider the extreme case in which every Wi and Wj have correlation coefficient ρi j = 1. In this case, our 20 measurements will be all the same and one measurement is as good as 20 measurements. To find the optimal value of c, we write a MATLAB function mquiz9(c) to calculate the MSE for a given c and second function that finds plots the MSE for a range of values of c. function [mse,af]=mquiz9(c); v1=ones(20,1); RW=toeplitz(c.ˆ((0:19)-1)); RY=(v1*(v1’)) +RW; af=(inv(RY))*v1; mse=1-((v1’)*af); function cmin=mquiz9minc(c); msec=zeros(size(c)); for k=1:length(c), [msec(k),af]=mquiz9(c(k)); end plot(c,msec); xlabel(’c’);ylabel(’e_Lˆ*’); [msemin,optk]=min(msec); cmin=c(optk); Note in mquiz9 that v1 corresponds to the vector 1 of all ones. The following commands finds the minimum c and also produces the following graph: >> c=0.01:0.01:0.99; >> mquiz9minc(c) ans = 0.4500 0 0.5 1 0.2 0.4 0.6 0.8 1 c e L * As we see in the graph, both small values and large values of c result in large MSE. 58
- 59. Quiz Solutions – Chapter 10 Quiz 10.1 There are many correct answers to this question. A correct answer specifies enough random variables to specify the sample path exactly. One choice for an alternate set of random variables that would specify m(t, s) is • m(0, s), the number of ongoing calls at the start of the experiment • N, the number of new calls that arrive during the experiment • X1, . . . , XN , the interarrival times of the N new arrivals • H, the number of calls that hang up during the experiment • D1, . . . , DH , the call completion times of the H calls that hang up Quiz 10.2 (1) We obtain a continuous time, continuous valued process when we record the temper- ature as a continuous waveform over time. (2) If at every moment in time, we round the temperature to the nearest degree, then we obtain a continuous time, discrete valued process. (3) If we sample the process in part (a) every T seconds, then we obtain a discrete time, continuous valued process. (4) Rounding the samples in part (c) to the nearest integer degree yields a discrete time, discrete valued process. Quiz 10.3 (1) Each resistor has resistance R in ohms with uniform PDF fR (r) = 0.01 950 ≤ r ≤ 1050 0 otherwise (1) The probability that a test produces a 1% resistor is p = P [990 ≤ R ≤ 1010] = 1010 990 (0.01) dr = 0.2 (2) 59
- 60. (2) In t seconds, exactly t resistors are tested. Each resistor is a 1% resistor with proba- bility p, independent of any other resistor. Consequently, the number of 1% resistors found has the binomial PMF PN(t) (n) = t n pn(1 − p)t−n n = 0, 1, . . . , t 0 otherwise (3) (3) First we will find the PMF of T1. This problem is easy if we view each resistor test as an independent trial. A success occurs on a trial with probability p if we find a 1% resistor. The first 1% resistor is found at time T1 = t if we observe failures on trials 1, . . . , t − 1 followed by a success on trial t. Hence, just as in Example 2.11, T1 has the geometric PMF PT1 (t) = (1 − p)t−1 p t = 1, 2, . . . 9 otherwise (4) Since p = 0.2, the probability the first 1% resistor is found in exactly five seconds is PT1(5) = (0.8)4(0.2) = 0.08192. (4) From Theorem 2.5, a geometric random variable with success probability p has ex- pected value 1/p. In this problem, E[T1] = 1/p = 5. (5) Note that once we find the first 1% resistor, the number of additional trials needed to find the second 1% resistor once again has a geometric PMF with expected value 1/p since each independent trial is a success with probability p. That is, T2 = T1 + T where T is independent and identically distributed to T1. Thus E [T2|T1 = 10] = E [T1|T1 = 10] + E T |T1 = 10 (5) = 10 + E T = 10 + 5 = 15 (6) Quiz 10.4 Since each Xi is a N(0, 1) random variable, each Xi has PDF fX(i) (x) = 1 √ 2π e−x2/2 (1) By Theorem 10.1, the joint PDF of X = X1 · · · Xn is fX (x) = fX(1),...,X(n) (x1, . . . , xn) = k i=1 fX (xi ) = 1 (2π)n/2 e−(x2 1+···+x2 n)/2 (2) 60
- 61. Quiz 10.5 The first and second hours are nonoverlapping intervals. Since one hour equals 3600 sec and the Poisson process has a rate of 10 packets/sec, the expected number of packets in each hour is E[Mi ] = α = 36, 000. This implies M1 and M2 are independent Poisson random variables each with PMF PMi (m) = αme−α m! m = 0, 1, 2, . . . 0 otherwise (1) Since M1 and M2 are independent, the joint PMF of M1 and M2 is PM1,M2 (m1, m2) = PM1 (m1) PM2 (m2) = ⎧ ⎪⎪⎨ ⎪⎪⎩ αm1+m2e−2α m1!m2! m1 = 0, 1, . . . ; m2 = 0, 1, . . . , 0 otherwise. (2) Quiz 10.6 To answer whether N (t) is a Poisson process, we look at the interarrival times. Let X1, X2, . . . denote the interarrival times of the N(t) process. Since we count only even- numbered arrival for N (t), the time until the first arrival of the N (t) is Y1 = X1 + X2. Since X1 and X2 are independent exponential (λ) random variables, Y1 is an Erlang (n = 2, λ) random variable; see Theorem 6.11. Since Yi (t), the ith interarrival time of the N (t) process, has the same PDF as Y1(t), we can conclude that the interarrival times of N (t) are not exponential random variables. Thus N (t) is not a Poisson process. Quiz 10.7 First, we note that for t > s, X(t) − X(s) = W(t) − W(s) √ α (1) Since W(t) − W(s) is a Gaussian random variable, Theorem 3.13 states that W(t) − W(s) is Gaussian with expected value E [X(t) − X(s)] = E [W(t) − W(s)] √ α = 0 (2) and variance E (W(t) − W(s))2 = E (W(t) − W(s))2 α = α(t − s) α (3) Consider s ≤ s < t. Since s ≥ s , W(t) − W(s) is independent of W(s ). This implies [W(t) − W(s)]/ √ α is independent of W(s )/ √ α for all s ≥ s . That is, X(t) − X(s) is independent of X(s ) for all s ≥ s . Thus X(t) is a Brownian motion process with variance Var[X(t)] = t. 61
- 62. Quiz 10.8 First we find the expected value µY (t) = µX (t) + µN (t) = µX (t). (1) To find the autocorrelation, we observe that since X(t) and N(t) are independent and since N(t) has zero expected value, E[X(t)N(t )] = E[X(t)]E[N(t )] = 0. Since RY (t, τ) = E[Y(t)Y(t + τ)], we have RY (t, τ) = E [(X(t) + N(t)) (X(t + τ) + N(t + τ))] (2) = E [X(t)X(t + τ)] + E [X(t)N(t + τ)] + E [X(t + τ)N(t)] + E [N(t)N(t + τ)] (3) = RX (t, τ) + RN (t, τ). (4) Quiz 10.9 From Definition 10.14, X1, X2, . . . is a stationary random sequence if for all sets of time instants n1, . . . , nm and time offset k, fXn1 ,...,Xnm (x1, . . . , xm) = fXn1+k,...,Xnm+k (x1, . . . , xm) (1) Since the random sequence is iid, fXn1 ,...,Xnm (x1, . . . , xm) = fX (x1) fX (x2) · · · fX (xm) (2) Similarly, for time instants n1 + k, . . . , nm + k, fXn1+k,...,Xnm+k (x1, . . . , xm) = fX (x1) fX (x2) · · · fX (xm) (3) We can conclude that the iid random sequence is stationary. Quiz 10.10 We must check whether each function R(τ) meets the conditions of Theorem 10.12: R(τ) ≥ 0 R(τ) = R(−τ) |R(τ)| ≤ R(0) (1) (1) R1(τ) = e−|τ| meets all three conditions and thus is valid. (2) R2(τ) = e−τ2 also is valid. (3) R3(τ) = e−τ cos τ is not valid because R3(−2π) = e2π cos 2π = e2π > 1 = R3(0) (2) (4) R4(τ) = e−τ2 sin τ also cannot be an autocorrelation function because R4(π/2) = e−π/2 sin π/2 = e−π/2 > 0 = R4(0) (3) 62
- 63. Quiz 10.11 (1) The autocorrelation of Y(t) is RY (t, τ) = E [Y(t)Y(t + τ)] (1) = E [X(−t)X(−t − τ)] (2) = RX (−t − (−t − τ)) = RX (τ) (3) Since E[Y(t)] = E[X(−t)] = µX , we can conclude that Y(t) is a wide sense stationary process. In fact, we see that by viewing a process backwards in time, we see the same second order statistics. (2) Since X(t) and Y(t) are both wide sense stationary processes, we can check whether they are jointly wide sense stationary by seeing if RXY (t, τ) is just a function of τ. In this case, RXY (t, τ) = E [X(t)Y(t + τ)] (4) = E [X(t)X(−t − τ)] (5) = RX (t − (−t − τ)) = RX (2t + τ) (6) Since RXY (t, τ) depends on both t and τ, we conclude that X(t) and Y(t) are not jointly wide sense stationary. To see why this is, suppose RX (τ) = e−|τ| so that samples of X(t) far apart in time have almost no correlation. In this case, as t gets larger, Y(t) = X(−t) and X(t) become less and less correlated. Quiz 10.12 From the problem statement, E [X(t)] = E [X(t + 1)] = 0 (1) E [X(t)X(t + 1)] = 1/2 (2) Var[X(t)] = Var[X(t + 1)] = 1 (3) The Gaussian random vector X = X(t) X(t + 1) has covariance matrix and corre- sponding inverse CX = 1 1/2 1/2 1 C−1 X = 4 3 1 −1/2 −1/2 1 (4) Since x C−1 X x = x0 x1 4 3 1 −1/2 −1/2 1 x0 x1 = 4 3 x2 0 − x0x+x2 1 (5) the joint PDF of X(t) and X(t + 1) is the Gaussian vector PDF fX(t),X(t+1) (x0, x1) = 1 (2π)n/2[det (CX)]1/2 exp − 1 2 x C−1 X x (6) = 1 √ 3π2 e−2 3 x2 0−x0x1+x2 1 (7) 63
- 64. 0 10 20 30 40 50 60 70 80 90 100 0 20 40 60 80 100 120 t M(t) Figure 4: Sample path of 100 minutes of the blocking switch of Quiz 10.13. Quiz 10.13 The simple structure of the switch simulation of Example 10.28 admits a deceptively simple solution in terms of the vector of arrivals A and the vector of departures D. With the introduction of call blocking. we cannot generate these vectors all at once. In particular, when an arrival occurs at time t, we need to know that M(t), the number of ongoing calls, satisfies M(t) < c = 120. Otherwise, when M(t) = c, we must block the call. Call blocking can be implemented by setting the service time of the call to zero so that the call departs as soon as it arrives. The blocking switch is an example of a discrete event system. The system evolves via a sequence of discrete events, namely arrivals and departures, at discrete time instances. A simulation of the system moves from one time instant to the next by maintaining a chrono- logical schedule of future events (arrivals and departures) to be executed. The program simply executes the event at the head of the schedule. The logic of such a simulation is 1. Start at time t = 0 with an empty system. Schedule the first arrival to occur at S1, an exponential (λ) random variable. 2. Examine the head-of-schedule event. • When the head-of-schedule event is the kth arrival is at time t, check the state M(t). – If M(t) < c, admit the arrival, increase the system state n by 1, and sched- ule a departure to occur at time t + Sn, where Sk is an exponential (λ) random variable. – If M(t) = c, block the arrival, do not schedule a departure event. • If the head of schedule event is a departure, reduce the system state n by 1. 3. Delete the head-of-schedule event and go to step 2. After the head-of-schedule event is completed and any new events (departures in this sys- tem) are scheduled, we know the system state cannot change until the next scheduled event. 64
- 65. Thus we know that M(t) will stay the same until then. In our simulation, we use the vector t as the set of time instances at which we inspect the system state. Thus for all times t(i) between the current head-of-schedule event and the next, we set m(i) to the current switch state. The complete program is shown in Figure 5. In most programming languages, it is common to implement the event schedule as a linked list where each item in the list has a data structure indicating an event timestamp and the type of the event. In MATLAB, a simple (but not elegant) way to do this is to have maintain two vectors: time is a list of timestamps of scheduled events and event is a the list of event types. In this case, event(i)=1 if the ith scheduled event is an arrival, or event(i)=-1 if the ith sched- uled event is a departure. When the program is passed a vector t, the output [m a b] is such that m(i) is the number of ongoing calls at time t(i) while a and b are the number of admits and blocks. The following instructions t=0:0.1:5000; [m,a,b]=simblockswitch(10,0.1,120,t); plot(t,m); generated a simulation lasting 5,000 minutes. A sample path of the first 100 minutes of that simulation is shown in Figure 4. The 5,000 minute full simulation produced a=49658 admitted calls and b=239 blocked calls. We can estimate the probability a call is blocked as ˆPb = b a + b = 0.0048. (1) In Chapter 12, we will learn that the exact blocking probability is given by Equation (12.93), a result known as the “Erlang-B formula.” From the Erlang-B formula, we can calculate that the exact blocking probability is Pb = 0.0057. One reason our simulation underesti- mates the blocking probability is that in a 5,000 minute simulation, roughly the first 100 minutes are needed to load up the switch since the switch is idle when the simulation starts at time t = 0. However, this says that roughly the first two percent of the simulation time was unusual. Thus this would account for only part of the disparity. The rest of the gap between 0.0048 and 0.0057 is that a simulation that includes only 239 blocks is not all that likely to give a very accurate result for the blocking probability. Note that in Chapter 12, we will learn that the blocking switch is an example of an M/M/c/c queue, a kind of Markov chain. Chapter 12 develops techniques for analyzing and simulating systems described by Markov chains that are much simpler than the discrete event simulation technique shown here. Nevertheless, for very complicated systems, the discrete event simulation is widely-used and often very efficient simulation method. 65
- 66. function [M,admits,blocks]=simblockswitch(lam,mu,c,t); blocks=0; %total # blocks admits=0; %total # admits M=zeros(size(t)); n=0; % # in system time=[ exponentialrv(lam,1) ]; event=[ 1 ]; %first event is an arrival timenow=0; tmax=max(t); while (timenow<tmax) M((timenow<=t)&(t<time(1)))=n; timenow=time(1); eventnow=event(1); event(1)=[ ]; time(1)= [ ]; % clear current event if (eventnow==1) % arrival arrival=timenow+exponentialrv(lam,1); % next arrival b4arrival=time<arrival; event=[event(b4arrival) 1 event(˜b4arrival)]; time=[time(b4arrival) arrival time(˜b4arrival)]; if n<c %call admitted admits=admits+1; n=n+1; depart=timenow+exponentialrv(mu,1); b4depart=time<depart; event=[event(b4depart) -1 event(˜b4depart)]; time=[time(b4depart) depart time(˜b4depart)]; else blocks=blocks+1; %one more block, immed departure disp(sprintf(’Time %10.3d Admits %10d Blocks %10d’,... timenow,admits,blocks)); end elseif (eventnow==-1) %departure n=n-1; end end Figure 5: Discrete event simulation of the blocking switch of Quiz 10.13. 66
- 67. Quiz Solutions – Chapter 11 Quiz 11.1 By Theorem 11.2, µY = µX ∞ −∞ h(t)dt = 2 ∞ 0 e−t dt = 2 (1) Since RX (τ) = δ(τ), the autocorrelation function of the output is RY (τ) = ∞ −∞ h(u) ∞ −∞ h(v)δ(τ + u − v) dv du = ∞ −∞ h(u)h(τ + u) du (2) For τ > 0, we have RY (τ) = ∞ 0 e−u e−τ−u du = e−τ ∞ 0 e−2u du = 1 2 e−τ (3) For τ < 0, we can deduce that RY (τ) = 1 2e−|τ| by symmetry. Just to be safe though, we can double check. For τ < 0, RY (τ) = ∞ −τ h(u)h(τ + u) du = ∞ −τ e−u e−τ−u du = 1 2 eτ (4) Hence, RY (τ) = 1 2 e−|τ| (5) Quiz 11.2 The expected value of the output is µY = µX ∞ n=−∞ hn = 0.5(1 + −1) = 0 (1) The autocorrelation of the output is RY [n] = 1 i=0 1 j=0 hi h j RX [n + i − j] (2) = 2RX [n] − RX [n − 1] − RX [n + 1] = 1 n = 0 0 otherwise (3) Since µY = 0, The variance of Yn is Var[Yn] = E[Y2 n ] = RY [0] = 1. 67
- 68. −15 −10 −5 0 5 10 15 0 0.2 0.4 0.6 f S X (f) −1500−1000 −500 0 500 1000 1500 0 2 4 6 8 x 10 f S X (f) −0.2 −0.1 0 0.1 0.2 −5 0 5 10 τ R X (τ) −2 −1 0 1 2 x 10 −3 −5 0 5 10 τ R X (τ) (a) W = 10 (b) W = 1000 Figure 6: The autocorrelation RX (τ) and power spectral density SX ( f ) for process X(t) in Quiz 11.5. Quiz 11.3 By Theorem 11.8, Y = Y33 Y34 Y35 is a Gaussian random vector since Xn is a Gaussian random process. Moreover, by Theorem 11.5, each Yn has expected value E[Yn] = µX ∞ n=−∞ hn = 0. Thus E[Y] = 0. Fo find the PDF of the Gaussian vector Y, we need to find the covariance matrix CY, which equals the correlation matrix RY since Y has zero expected value. One way to find the RY is to observe that RY has the Toeplitz structure of Theorem 11.6 and to use Theorem 11.5 to find the autocorrelation function RY [n] = ∞ i=−∞ ∞ j=−∞ hi h j RX [n + i − j]. (1) Despite the fact that RX [k] is an impulse, using Equation (1) is surprisingly tedious because we still need to sum over all i and j such that n + i − j = 0. In this problem, it is simpler to observe that Y = HX where X = X30 X31 X32 X33 X34 X35 (2) and H = 1 4 ⎡ ⎣ 1 1 1 1 0 0 0 1 1 1 1 0 0 0 1 1 1 1 ⎤ ⎦ . (3) In this case, following Theorem 11.7, or by directly applying Theorem 5.13 with µX = 0 and A = H, we obtain RY = HRXH . Since RX [n] = δn, RX = I, the identity matrix. 68
- 69. Thus CY = RY = HH = 1 16 ⎡ ⎣ 4 3 2 3 4 3 2 3 4 ⎤ ⎦ . (4) It follows (very quickly if you use MATLAB for 3 × 3 matrix inversion) that C−1 Y = 16 ⎡ ⎣ 7/12 −1/2 1/12 −1/2 1 −1/2 1/12 −1/2 7/12 ⎤ ⎦ . (5) Thus, the PDF of Y is fY (y) = 1 (2π)3/2[det (CY)]1/2 exp − 1 2 y C−1 Y y . (6) A disagreeable amount of algebra will show det(CY) = 3/1024 and that the PDF can be “simplified” to fY (y) = 16 √ 6π3 exp −8 7 12 y2 33 + y2 34 + 7 12 y2 35 − y33y34 + 1 6 y33y35 − y34y35 . (7) Equation (7) shows that one of the nicest features of the multivariate Gaussian distribution is that y C−1 Y y is a very concise representation of the cross-terms in the exponent of fY(y). Quiz 11.4 This quiz is solved using Theorem 11.9 for the case of k = 1 and M = 2. In this case, Xn = Xn−1 Xn and RXn = RX [0] RX [1] RX [1] RX [0] = 1.1 0.9 0.9 1.1 (1) and RXn Xn+1 = E Xn−1 Xn Xn+1 = RX [2] RX [1] = 0.81 0.9 . (2) The MMSE linear first order filter for predicting Xn+1 at time n is the filter h such that ←− h = R−1 Xn RXn Xn+1 = 1.1 0.9 0.9 1.1 −1 0.81 0.9 = 1 400 81 261 . (3) It follows that the filter is h = 261/400 81/400 and the MMSE linear predictor is ˆXn+1 = 81 400 Xn−1 + 261 400 Xn. (4) to find the mean square error, one approach is to follow the method of Example 11.13 and to directly calculate e∗ L = E (Xn+1 − ˆXn+1)2 . (5) 69
- 70. This method is workable for this simple problem but becomes increasingly tedious for higher order filters. Instead, we can derive the mean square error for an arbitary prediction filter h. Since ˆXn+1 = ←− h Xn, e∗ L = E Xn+1 − ←− h Xn 2 (6) = E (Xn+1 − ←− h Xn)(Xn+1 − ←− h Xn) (7) = E (Xn+1 − ←− h Xn)(Xn+1 − Xn ←− h ) (8) After a bit of algebra, we obtain e∗ L = RX [0] − 2 ←− h RXn Xn+1 + ←− h RXn ←− h (9) (10) with the substitution ←− h = R−1 Xn RXn Xn+1, we obtain e∗ L = RX [0] − RXn Xn+1 R−1 Xn RXn Xn+1 (11) = RX [0] − ←− h RXn Xn+1 (12) Note that this is essentially the same result as Theorem 9.7 with Y = Xn, X = Xn+1 and ˆa = ←− h . It is noteworthy that the result is derived in a much simpler way in the proof of Theorem 9.7 by using the orthoginality property of the LMSE estimator. In any case, the mean square error is e∗ L = RX [0] − ←− h RXn Xn+1 = 1.1 − 1 400 81 261 0.81 0.9 = 506 1451 = 0.3487. (13) recalling that the blind estimate would yield a mean square error of Var[X] = 1.1, we see that observing Xn−1 and Xn improves the accuracy of our prediction of Xn+1. Quiz 11.5 (1) By Theorem 11.13(b), the average power of X(t) is E X2 (t) = ∞ −∞ SX ( f ) d f = W −W 5 W d f = 10 Watts (1) (2) The autocorrelation function is the inverse Fourier transform of SX ( f ). Consulting Table 11.1, we note that SX ( f ) = 10 1 2W rect f 2W (2) It follows that the inverse transform of SX ( f ) is RX (τ) = 10 sinc(2Wτ) = 10 sin(2πWτ) 2πWτ (3) (3) For W = 10 Hz and W = 1 kHZ, graphs of SX ( f ) and RX (τ) appear in Figure 6. 70
- 71. Quiz 11.6 In a sampled system, the discrete time impulse δ[n] has a flat discrete Fourier transform. That is, if RX [n] = 10δ[n], then SX (φ) = ∞ n=−∞ 10δ[n]e− j2πφn = 10 (1) Thus, RX [n] = 10δ[n]. (This quiz is really lame!) Quiz 11.7 Since Y(t) = X(t − t0), RXY (t, τ) = E [X(t)Y(t + τ)] = E [X(t)X(t + τ − t0)] = RX (τ − t0) (1) We see that RXY (t, τ) = RXY (τ) = RX (τ − t0). From Table 11.1, we recall the prop- erty that g(τ − τ0) has Fourier transform G( f )e− j2π f τ0. Thus the Fourier transform of RXY (τ) = RX (τ − t0) = g(τ − t0) is SXY ( f ) = SX ( f )e− j2π f t0. (2) Quiz 11.8 We solve this quiz using Theorem 11.17. First we need some preliminary facts. Let a0 = 5,000 so that RX (τ) = 1 a0 a0e−a0|τ| . (1) Consulting with the Fourier transforms in Table 11.1, we see that SX ( f ) = 1 a0 2a2 0 a2 0 + (2π f )2 = 2a0 a2 0 + (2π f )2 (2) The RC filter has impulse response h(t) = a1e−a1tu(t), where u(t) is the unit step function and a1 = 1/RC where RC = 10−4 is the filter time constant. From Table 11.1, H( f ) = a1 a1 + j2π f (3) (1) Theorem 11.17, SXY ( f ) = H( f )SX ( f ) = 2a0a1 [a1 + j2π f ] a2 0 + (2π f )2 . (4) (2) Again by Theorem 11.17, SY ( f ) = H∗ ( f )SXY ( f ) = |H( f )|2 SX ( f ). (5) 71
- 72. Note that |H( f )|2 = H( f )H∗ ( f ) = a1 (a1 + j2π f ) a1 (a1 − j2π f ) = a2 1 a2 1 + (2π f )2 (6) Thus, SY ( f ) = |H( f )|2 SX ( f ) = 2a0a2 1 a2 1 + (2π f )2 a2 0 + (2π f )2 (7) (3) To find the average power at the filter output, we can either use basic calculus and calculate ∞ −∞ SY ( f ) d f directly or we can find RY (τ) as an inverse transform of SY ( f ). Using partial fractions and the Fourier transform table, the latter method is actually less algebra. In particular, some algebra will show that SY ( f ) = K0 a2 0 + (2π f )2 + K1 a1 + (2π f )2 (8) where K0 = 2a0a2 1 a2 1 − a2 0 , K1 = −2a0a2 1 a2 1 − a2 0 . (9) Thus, SY ( f ) = K0 2a2 0 2a2 0 a2 0 + (2π f )2 + K1 2a2 1 2a2 1 a1 + (2π f )2 . (10) Consulting with Table 11.1, we see that RY (τ) = K0 2a2 0 a0e−a0|τ| + K1 2a2 1 a1e−a1|τ| (11) Substituting the values of K0 and K1, we obtain RY (τ) = a2 1e−a0|τ| − a0a1e−a1|τ| a2 1 − a2 0 . (12) The average power of the Y(t) process is RY (0) = a1 a1 + a0 = 2 3 . (13) Note that the input signal has average power RX (0) = 1. Since the RC filter has a 3dB bandwidth of 10,000 rad/sec and the signal X(t) has most of its its signal energy below 5,000 rad/sec, the output signal has almost as much power as the input. 72
- 73. Quiz 11.9 This quiz implements an example of Equations (11.146) and (11.147) for a system in which we filter Y(t) = X(t) + N(t) to produce an optimal linear estimate of X(t). The solution to this quiz is just to find the filter ˆH( f ) using Equation (11.146) and to calculate the mean square error eL∗ using Equation (11.147). Comment: Since the text omitted the derivations of Equations (11.146) and (11.147), we note that Example 10.24 showed that RY (τ) = RX (τ) + RN (τ), RY X (τ) = RX (τ). (1) Taking Fourier transforms, it follows that SY ( f ) = SX ( f ) + SN ( f ), SY X ( f ) = SX ( f ). (2) Now we can go on to the quiz, at peace with the derivations. (1) Since µN = 0, RN (0) = Var[N] = 1. This implies RN (0) = ∞ −∞ SN ( f ) d f = B −B N0 d f = 2N0 B (3) Thus N0 = 1/(2B). Because the noise process N(t) has constant power RN (0) = 1, decreasing the single-sided bandwidth B increases the power spectral density of the noise over frequencies | f | < B. (2) Since RX (τ) = sinc(2Wτ), where W = 5,000 Hz, we see from Table 11.1 that SX ( f ) = 1 104 rect f 104 . (4) The noise power spectral density can be written as SN ( f ) = N0 rect f 2B = 1 2B rect f 2B , (5) From Equation (11.146), the optimal filter is ˆH( f ) = SX ( f ) SX ( f ) + SN ( f ) = 1 104 rect f 104 1 104 rect f 104 + 1 2B rect f 2B . (6) 73
- 74. (3) We produce the output ˆX(t) by passing the noisy signal Y(t) through the filter ˆH( f ). From Equation (11.147), the mean square error of the estimate is e∗ L = ∞ −∞ SX ( f )SN ( f ) SX ( f ) + SN ( f ) d f (7) = ∞ −∞ 1 104 rect f 104 1 2B rect f 2B 1 104 rect f 104 + 1 2B rect f 2B d f. (8) To evaluate the MSE e∗ L, we need to whether B ≤ W. Since the problem asks us to find the largest possible B, let’s suppose B ≤ W. We can go back and consider the case B > W later. When B ≤ W, the MSE is e∗ L = B −B 1 104 1 2B 1 104 + 1 2B d f = 1 104 1 104 + 1 2B = 1 1 + 5,000 B (9) To obtain MSE e∗ L ≤ 0.05 requires B ≤ 5,000/19 = 263.16 Hz. Although this completes the solution to the quiz, what is happening may not be obvious. The noise power is always Var[N] = 1 Watt, for all values of B. As B is decreased, the PSD SN ( f ) becomes increasingly tall, but only over a bandwidth B that is decreasing. Thus as B descreases, the filter ˆH( f ) makes an increasingly deep and narrow notch at frequencies | f | ≤ B. Two examples of the filter ˆH( f ) are shown in Figure 7. As B shrinks, the filter suppresses less of the signal of X(t). The result is that the MSE goes down. Finally, we note that we can choose B very large and also achieve MSE e∗ L = 0.05. In particular, when B > W = 5000, SN ( f ) = 1/2B over frequencies | f | < W. In this case, the Wiener filter ˆH( f ) is an ideal (flat) lowpass filter ˆH( f ) = ⎧ ⎨ ⎩ 1 104 1 104 + 1 2B | f | < 5,000, 0 otherwise. (10) Thus increasing B spreads the constant 1 watt of power of N(t) over more bandwidth. The Wiener filter removes the noise that is outside the band of the desired signal. The mean square error is e∗ L = 5000 −5000 1 104 1 2B 1 104 + 1 2B d f = 1 2B 1 104 + 1 2B = 1 B 5000 + 1 (11) In this case, B ≥ 9.5 × 104 guarantees e∗ L ≤ 0.05. Quiz 11.10 It is fairly straightforward to find SX (φ) and SY (φ). The only thing to keep in mind is to use fftc to transform the autocorrelation RX [ f ] into the power spectral density SX (φ). The following MATLAB program generates and plots the functions shown in Figure 8 74
- 75. −5000 −2000 0 2000 5000 0 0.5 1 f H(f) −5000 −2000 0 2000 5000 0 0.5 1 f H(f) B = 500 B = 2500 Figure 7: Wiener filter for Quiz 11.9. %mquiz11.m N=32; rx=[2 4 2]; SX=fftc(rx,N); %autocorrelation and PSD stem(0:N-1,abs(sx)); xlabel(’n’);ylabel(’S_X(n/N)’); h2=0.5*[1 1]; H2=fft(h2,N); %impulse/filter response: M=2 SY2=SX.* ((abs(H2)).ˆ2); figure; stem(0:N-1,abs(SY2)); %PSD of Y for M=2 xlabel(’n’);ylabel(’S_{Y_2}(n/N)’); h10=0.1*ones(1,10); H10=fft(h10,N); %impulse/filter response: M=10 SY10=sx.*((abs(H10)).ˆ2); figure; stem(0:N-1,abs(SY10)); xlabel(’n’);ylabel(’S_{Y_{10}}(n/N)’); Relative to M = 2, when M = 10, the filter H(φ) filters out almost all of the high frequency components of X(t). In the context of Example 11.26, the low pass moving average filter for M = 10 removes the high frquency components and results in a filter output that varies very slowly. As an aside, note that the vectors SX, SY2 and SY10 in mquiz11 should all be real- valued vectors. However, the finite numerical precision of MATLAB results in tiny imagi- nary parts. Although these imaginary parts have no computational significance, they tend to confuse the stem function. Hence, we generate stem plots of the magnitude of each power spectral density. 75
- 76. 0 5 10 15 20 25 30 35 0 5 10 n S X (n/N) 0 5 10 15 20 25 30 35 0 5 10 n S Y 2 (n/N) 0 5 10 15 20 25 30 35 0 5 10 n S Y 10 (n/N) Figure 8: For Quiz 11.10, graphs of SX (φ), SY (n/N) for M = 2, and Sφ(n/N) for M = 10 using an N = 32 point DFT. 76
- 77. Quiz Solutions – Chapter 12 Quiz 12.1 The system has two states depending on whether the previous packet was received in error. From the problem statement, we are given the conditional probabilities P Xn+1 = 0|Xn = 0 = 0.99 P Xn+1 = 1|Xn = 1 = 0.9 (1) Since each Xn must be either 0 or 1, we can conclude that P Xn+1 = 1|Xn = 0 = 0.01 P Xn+1 = 0|Xn = 1 = 0.1 (2) These conditional probabilities correspond to the transition matrix and Markov chain: 0 1 0.01 0.1 0.99 0.9 P = 0.99 0.01 0.10 0.90 (3) Quiz 12.2 From the problem statement, the Markov chain and the transition matrix are 0 1 1 0.6 0.2 0.2 0.6 0.4 0.6 0.4 P = ⎡ ⎣ 0.4 0.6 0 0.2 0.6 0.2 0 0.6 0.4 ⎤ ⎦ (1) The eigenvalues of P are λ1 = 0 λ2 = 0.4 λ3 = 1 (2) We can diagonalize P into P = S−1 DS = ⎡ ⎣ −0.6 0.5 1 0.4 0 1 −0.6 −0.5 1 ⎤ ⎦ ⎡ ⎣ λ1 0 0 0 λ2 0 0 0 λ3 ⎤ ⎦ ⎡ ⎣ −0.5 1 −0.5 1 0 −1 0.2 0.6 0.2 ⎤ ⎦ (3) where si , the ith row of S, is the left eigenvector of P satisfying si P = λi si . Algebra will verify that the n-step transition matrix is Pn = S−1 Dn S = ⎡ ⎣ 0.2 0.6 0.2 0.2 0.6 0.2 0.2 0.6 0.2 ⎤ ⎦ + (0.4)n ⎡ ⎣ 0.5 0 −0.5 0 0 0 −0.5 0 0.5 ⎤ ⎦ (4) Quiz 12.3 The Markov chain describing the factory status and the corresponding state transition matrix are 77
- 78. 2 0 1 0.9 0.1 11 P = ⎡ ⎣ 0.9 0.1 0 0 0 1 1 0 0 ⎤ ⎦ (1) With π = π0 π1 π2 , the system of equations π = π P yields π1 = 0.1π0 and π2 = π1. This implies π0 + π1 + π2 = π0(1 + 0.1 + 0.1) = 1 (2) It follows that the limiting state probabilities are π0 = 5/6, π1 = 1/12, π2 = 1/12. (3) Quiz 12.4 The communicating classes are C1 = {0, 1} C2 = {2, 3} C3 = {4, 5, 6} (1) The states in C1 and C3 are aperiodic. The states in C2 have period 2. Once the system enters a state in C1, the class C1 is never left. Thus the states in C1 are recurrent. That is, C1 is a recurrent class. Similarly, the states in C3 are recurrent. On the other hand, the states in C2 are transient. Once the system exits C2, the states in C2 are never reentered. Quiz 12.5 At any time t, the state n can take on the values 0, 1, 2, . . .. The state transition proba- bilities are Pn−1,n = P [K > n|K > n − 1] = P [K > n] P [K > n − 1] (1) Pn−1,0 = P [K = n|K > n − 1] = P [K = n] P [K > n − 1] (2) (3) The Markov chain resembles 0 1 P K=2[ ] P K=[ 1] 3 4 P K=4[ ] 2 P K=3[ ] P K=5[ ] 1 1 1 1 1 …... 78
- 79. The stationary probabilities satisfy π0 = π0 P [K = 1] + π1, (4) π1 = π0 P [K = 2] + π2, (5) ... πk−1 = π0 P [K = k] + πk, k = 1, 2, . . . (6) From Equation (4), we obtain π1 = π0 (1 − P [K = 1]) = π0 P [K > 1] (7) Similarly, Equation (5) implies π2 = π1 − π0 P [K = 2] = π0 (P [K > 1] − P [K = 2]) = π0 P [K > 2] (8) This suggests that πk = π0 P[K > k]. We verify this pattern by showing that πk = π0 P[K > k] satisfies Equation (6): π0 P [K > k − 1] = π0 P [K = k] + π0 P [K > k] . (9) When we apply ∞ k=0 πk = 1, we obtain π0 ∞ n=0 P[K > k] = 1. From Problem 2.5.11, we recall that ∞ k=0 P[K > k] = E[K]. This implies πn = P [K > n] E [K] (10) This Markov chain models repeated random countdowns. The system state is the time until the counter expires. When the counter expires, the system is in state 0, and we randomly reset the counter to a new value K = k and then we count down k units of time. Since we spend one unit of time in each state, including state 0, we have k − 1 units of time left after the state 0 counter reset. If we have a random variable W such that the PMF of W satisfies PW (n) = πn, then W has a discrete PMF representing the remaining time of the counter at a time in the distant future. Quiz 12.6 (1) By inspection, the number of transitions need to return to state 0 is always a multiple of 2. Thus the period of state 0 is d = 2. (2) To find the stationary probabilities, we solve the system of equations π = πP and 3 i=0 πi = 1: π0 = (3/4)π1 + (1/4)π3 (1) π1 = (1/4)π0 + (1/4)π2 (2) π2 = (1/4)π1 + (3/4)π3 (3) 1 = π0 + π1 + π2 + π3 (4) 79
- 80. Solving the second and third equations for π2 and π3 yields π2 = 4π1 − π0 π3 = (4/3)π2 − (1/3)π1 = 5π1 − (4/3)π0 (5) Substituting π3 back into the first equation yields π0 = (3/4)π1 + (1/4)π3 = (3/4)π1 + (5/4)π1 − (1/3)π0 (6) This implies π1 = (2/3)π0. It follows from the first and second equations that π2 = (5/3)π0 and π3 = 2π0. Lastly, we choose π0 so the state probabilities sum to 1: 1 = π0 + π1 + π2 + π3 = π0 1 + 2 3 + 5 3 + 2 = 16 3 π0 (7) It follows that the state probabilities are π0 = 3 16 π1 = 2 16 π2 = 5 16 π3 = 6 16 (8) (3) Since the system starts in state 0 at time 0, we can use Theorem 12.14 to find the limiting probability that the system is in state 0 at time nd: lim n→∞ P00(nd) = dπ0 = 3 8 (9) Quiz 12.7 The Markov chain has the same structure as that in Example 12.22. The only difference is the modified transition rates: 0 1 1 3 4 ( )2/3 a 1 - ( )2/3 a ( )3/4 a 1 - 3/4( ) a ( )4/5 a 1 - 4/5( )a 2 ( )1/2 a 1- 1/2( ) a … The event T00 > n occurs if the system reaches state n before returning to state 0, which occurs with probability P [T00 > n] = 1 × 1 2 α × 2 3 α × · · · × n − 1 n α = 1 n α . (1) Thus the CDF of T00 satisfies FT00(n) = 1−P[T00 > n] = 1−1/nα. To determine whether state 0 is recurrent, we observe that for all α > 0 P [V00] = lim n→∞ FT00 (n) = lim n→∞ 1 − 1 nα = 1. (2) 80
- 81. Thus state 0 is recurrent for all α > 0. Since the chain has only one communicating class, all states are recurrent. ( We also note that if α = 0, then all states are transient.) To determine whether the chain is null recurrent or positive recurrent, we need to calcu- late E[T00]. In Example 12.24, we did this by deriving the PMF PT00(n). In this problem, it will be simpler to use the result of Problem 2.5.11 which says that ∞ k=0 P[K > k] = E[K] for any non-negative integer-valued random variable K. Applying this result, the expected time to return to state 0 is E [T00] = ∞ n=0 P [T00 > n] = 1 + ∞ n=1 1 nα . (3) For 0 < α ≤ 1, 1/nα ≥ 1/n and it follows that E [T00] ≥ 1 + ∞ n=1 1 n = ∞. (4) We conclude that the Markov chain is null recurrent for 0 < α ≤ 1. On the other hand, for α > 1, E [T00] = 2 + ∞ n=2 1 nα . (5) Note that for all n ≥ 2 1 nα ≤ n n−1 dx xα (6) This implies E [T00] ≤ 2 + ∞ n=2 n n−1 dx xα (7) = 2 + ∞ 1 dx xα (8) = 2 + x−α+1 −α + 1 ∞ 1 = 2 + 1 α − 1 < ∞ (9) Thus for all α > 1, the Markov chain is positive recurrent. Quiz 12.8 The number of customers in the ”friendly” store is given by the Markov chain 1 i i+1 p p p ( )( )1-p 1-q ( )( )1-p 1-q ( )( )1-p 1-q ( )( )1-p 1-q ( )1-p q ( )1-p q ( )1-p q ( )1-p q 0 ××× ××× 81
- 82. In the above chain, we note that (1 − p)q is the probability that no new customer arrives, an existing customer gets one unit of service and then departs the store. By applying Theorem 12.13 with state space partitioned between S = {0, 1, . . . , i} and S = {i + 1, i + 2, . . .}, we see that for any state i ≥ 0, πi p = πi+1(1 − p)q. (1) This implies πi+1 = p (1 − p)q πi . (2) Since Equation (2) holds for i = 0, 1, . . ., we have that πi = π0αi where α = p (1 − p)q . (3) Requiring the state probabilities to sum to 1, we have that for α < 1, ∞ i=0 πi = π0 ∞ i=0 αi = π0 1 − α = 1. (4) Thus for α < 1, the limiting state probabilities are πi = (1 − α)αi , i = 0, 1, 2, . . . (5) In addition, for α ≥ 1 or, equivalently, p ≥ q/(1 − q), the limiting state probabilities do not exist. Quiz 12.9 The continuous time Markov chain describing the processor is 0 1 2 3.01 3 4 2 3 2 3 2 2 3 0.01 0.01 0.01 Note that q10 = 3.1 since the task completes at rate 3 per msec and the processor reboots at rate 0.1 per msec and the rate to state 0 is the sum of those two rates. From the Markov chain, we obtain the following useful equations for the stationary distribution. 5.01p1 = 2p0 + 3p2 5.01p2 = 2p1 + 3p3 5.01p3 = 2p2 + 3p4 3.01p4 = 2p3 We can solve these equations by working backward and solving for p4 in terms of p3, p3 in terms of p2 and so on, yielding p4 = 20 31 p3 p3 = 620 981 p2 p2 = 19620 31431 p1 p1 = 628, 620 1, 014, 381 p0 (1) 82
- 83. Applying p0 + p1 + p2 + p3 + p4 = 1 yields p0 = 1, 014, 381/2, 443, 401 and the stationary probabilities are p0 = 0.4151 p1 = 0.2573 p2 = 0.1606 p3 = 0.1015 p4 = 0.0655 (2) Quiz 12.10 The M/M/c/∞ queue has Markov chain c c+110 λ λ λ λ λ µ 2µ cµ cµ cµ From the Markov chain, the stationary probabilities must satisfy pn = (ρ/n)pn−1 n = 1, 2, . . . , c (ρ/c)pn−1 n = c + 1, c + 2, . . . (1) It is straightforward to show that this implies pn = p0ρn/n! n = 1, 2, . . . , c p0 (ρ/c)n−c ρc/c! n = c + 1, c + 2, . . . (2) The requirement that ∞ n=0 pn = 1 yields p0 = c n=0 ρn /n! + ρc c! ρ/c 1 − ρ/c −1 (3) 83